2 History of Math for the Liberal Arts CHAPTER 6 · PDF file41 The Slide Rule ... Aryabhata advanced trigonometry, ... 183 represents the ones place, we add up all numbers in that

2001, Lawrence Morales; MAT107 Chapter 6 – Page 1

1

2

History of Math 3

for the Liberal Arts 4

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CHAPTER 6 6

7

Two Great Achievements: 8

Logarithms & Cubic 9

Equations 10

11

12

Lawrence Morales 13

Seattle Central 14

Community College 15


TABLE OF CONTENTS 16

TABLE OF CONTENTS ............................................................................................................. 2 17

PART 1: Introduction and Non-Western Math, and Evolving Calculating Methods............ 4 18

Historical Background................................................................................................................ 4 19

Multiplication from India............................................................................................................ 5 20

Gelosia Multiplication ................................................................................................................ 6 21

PART 2: The Solution of the Cubic and Quartic Equations .................................................. 10 22

Historical and Mathematical Background................................................................................ 10 23

Cardano and His Gang of Italian Algebraists.......................................................................... 11 24

Ars Magna and the Solution of the Cubic................................................................................. 13 25

Solving the General Cubic Equation ........................................................................................ 18 26

The Solution of the Quartic Equation and Beyond ................................................................... 21 27

PART 3: Modern Calculations .................................................................................................. 22 28

Decimal Fractions .................................................................................................................... 22 29

PART 4: Napier and The Emergence of Logarithms.............................................................. 25 30

The Idea and Use of Logarithms............................................................................................... 27 31

The Idea Behind Logs ............................................................................................................... 27 32

Rule of Logs…One Step Closer to the Tables........................................................................... 28 33

The Log Tables.......................................................................................................................... 34 34

Larger Numbers and the Log Tables ........................................................................................ 36 35

Logs of Very Large Numbers .................................................................................................... 38 36

The Antilog Tables .................................................................................................................... 38 37

The Evil Twins Meet Each Other…Finally............................................................................... 41 38

Complex Calculations............................................................................................................... 41 39

PART 5: New Calculating Devices ............................................................................................ 46 40

The Slide Rule ........................................................................................................................... 46 41

Napier and Other Calculating Devices..................................................................................... 46 42

PART 6: Appendix A − Solution of the Quartic ...................................................................... 49 43

PART 7: Appendix B – Log and Antilog Tables...................................................................... 51 44

PART 8: Appendix C – Napier’s Rods ..................................................................................... 55 45


PART 9: Homework ................................................................................................................... 57 46

Multiplication with the Gelosia Grid System............................................................................ 57 47

Checking Solutions of Cubic Equations.................................................................................... 57 48

Using Cardano’s Formula on Depressed Cubics..................................................................... 57 49

Using Cardano’s Formula on Non-Depressed Cubics............................................................. 58 50

Stevin’s Notation....................................................................................................................... 58 51

Basic Logarithm Rules.............................................................................................................. 59 52

Using the Log Tables ................................................................................................................ 60 53

Using the Antilog Tables........................................................................................................... 60 54

Using Log and Antilog Tables to Do Calculations................................................................... 60 55

Applications of Logarithms....................................................................................................... 61 56

Writing ...................................................................................................................................... 62 57

Blank Gelosia Grid ......................................................................Error! Bookmark not defined. 58

PART 10: Endnotes .................................................................................................................... 63 59


PART 1: Introduction and Non-Western Math, and Evolving 60 Calculating Methods 61 62 Historical Background 63 64 After the Greeks, a variety of cultures spread previous mathematical accomplishments and also 65 created many of their own. For example, in the last three centuries B.C.E., China worked on 66 ideas of square and cube roots as well as methods of solving systems of linear equations. They 67 also tackled mathematical surveying techniques during this period. In Egypt, Apollonius 68 developed a theory of conic sections. From 0 to 400 C.E., Ptolemy made inroads into astronomy 69 while Hypatia, the famous female mathematician, made a name for herself by commenting and 70 lecturing on the work of Apollonius and by being a gifted and inspiring teacher of mathematics 71 and philosophy. 72 73 From 400−800, the Italian, Boethius, contributed to the field by writing arithmetic books that 74 were used by Europe during a time when mathematics in that part of the world was in decline. 75 His book, Arithmetic taught others about Pythagorean number theory. On the other side of the 76 world, the Mayan numeration system was developed and used for astronomical purposes. And in 77 India, a major mathematical movement emerged: Aryabhata advanced trigonometry, while 78 Brahmagupta made major contributions in mathematics and astronomy. It was also during this 79 time that the Hindu−Arabic decimal place−value number system began to emerge and gain 80 popularity. 81 82 From 800−1000, India continued its work as it developed algebraic techniques. In what is 83 modern−day Iraq, Al−Khwarizmi wrote an influential text on algebra whose title actually gives 84 us the word “algebra.” (The title was “Hisab al-jabr w’al-muqabala.”) Other Islamic 85 mathematicians were also hard at work, not only preserving and translating ancient Greek texts, 86 but making their own advances, especially in the area of algebra. During this time, Spain became 87 a passageway for the Hindu−Arabic numbers into Europe. 88 89 From 1000−1200, Islamic mathematics continued to develop with work on what is now known 90 as Pascal’s triangle, found geometric solutions to certain cubic equations (important in this 91 chapter), and explore sums of powers. In India, Al−Biruni advanced spherical trigonometry, 92 while in China, Pascal’s triangle (as it is now called) was used to solve equations. In Spain, 93 Arabic works were translated into Latin, which would be important in the coming resurgence in 94 European mathematics. Toward that end, Leonardo of Pisa advocated the use of Hindu−Arabic 95 numbers. In Italy, the rich world of Islamic mathematics was introduced and began to spur 96 interest. 97 98 From 1200−1600, major contributions continued to flow from the Islamic world, China, and 99 India. In England, new algebra and trigonometry texts emerged. In France, Viète pushed a new 100 decimal fraction system. During this time, Copernicus proposed a new heliocentric theory that 101 would greatly affect how mathematics developed from that time onwards. In Italy, a group called 102 the “Italian Algebraists” conquered the problem of finding an equation for solving cubic 103 (third−degree) and quartic (fourth−degree) polynomial equations, which we will study in more 104 detail later in this chapter. 105


Many of these intermediate topics are worthy of their own independent treatment. Due to time 106 considerations, we will look at just one interesting contribution by Bhaskara. (Future versions of 107 this text will have more information on Islamic, Hindu, and Chinese mathematics). Later, we will 108 look at two major achievements in the history of mathematics. The first topic we will explore in 109 more detail is the algebraic solution of cubic and quartic equations. The second achievement we 110 will examine is the set of tools that were developed which greatly simplified the process of doing 111 long, complex computations. These tools would enable mathematicians and scientists to make 112 great inroads into their fields of studies during the 1600’s and 1700’s. Specifically, we will 113 focus on the invention of logarithms by John Napier and the dramatic impact they had on the 114 mathematical landscape. 115 116 Multiplication from India 117 118 In the 12th century, the Indian mathematician Bhaskara (also known as Bhasharacharaya in India) 119 represented the peak of mathematical knowledge. He was the head of an astronomy observatory 120 at Ujjain, which was the prominent center for mathematics in India at the time. He had a 121 thorough understanding of the number 0, negative numbers, and methods of solving equations − 122 centuries ahead of the Europeans. He established his reputation (in part) based on a work titled 123 Lilavati (“The Beautiful”), which had 13 chapters and covered topics in arithmetic, geometry, 124 and algebra. 125 126 One of the interesting 127 contributions he 128 included in his work was 129 a proof of the 130 Pythagorean theorem. 131 He started with a square 132 that was cut into four 133 triangles and one square, 134 and then rearranged 135 them to create two 136 squares that were 137 positioned immediately 138 next to each other. His 139 picture was accompanied by the simple phrase, “Behold!” 140 141

Think About It

Why is this a proof of the Pythagorean

Theorem?


142 This appears to have been inspired by the Chinese and 143 one of their own “proofs” of the Pythagorean Theorem. 144 Recall the hsuan−thu from the Pythagorean chapter. 145 You can see the corresponding inner−squares embedded 146 in the middle of the larger squares with side c. This 147 smaller square gets moved into the upper right hand 148 corner of the two larger squares that are positioned side 149 by side, above. 150 151 Gelosia Multiplication 152 153 In Lilavati, Bhaskara also provides five different 154 methods for multiplication. One of these is interesting 155 because it later emerged in the Middle Ages and was eventually adapted by the inventor of 156 logarithms, John Napier. It is now called the gelosia method of multiplication. The word gelosia 157 means “lattice” or “grating.” 158 159 In this method a grid is set up, several simple multiplications are placed into the grid, and then 160

diagonals on the grid are added together to get a final result. For 161 example, to do 286×734, a three−by−three grid would be drawn 162 with the two numbers to be multiplied written on the top and right 163 side. 164 165 From the first picture (left), which looks a lot like a grid or lattice, 166 you can see where the method gets its name. To fill in the grid, you 167 take the number in a column and the number in a row and multiply 168 them. You place the result in the square where that row and column 169 intersect each other. For example, column 8 and row 3 give a 170 product of 24, so 24 goes in that square. Note that the tens digit 171 goes to the left of the diagonal of that “cell” 172 while the ones goes to the right of the 173 diagonal. If the product of a column and 174 row is less than ten, we put a 0 in the tens 175 position…all spaces should be filled in to 176 avoid confusion. The picture (left) shows 177 all the cells filled in. 178 179 This method allows you to do all the 180

multiplication at once…it’s only after this is done that we move to 181 addition. The addition is done along diagonals. Starting in the lower right corner, which 182 represents the ones place, we add up all numbers in that diagonal. If we get more than ten for the 183 sum, we carry any groups of ten into the next diagonal up. Hence, in the second diagonal, we 184 have 8+2+2=12, so we carry 1 up to the next diagonal and keep the 2 (just like when we add 185 vertically). The third diagonal has a sum of 1+8+3+4+1+2= 19. Again, that 1 will carry up and 186 the 9 is kept and recorded for that row. Continuing in this manner will produce a digit for each 187

2 4


diagonal and we can read the final answer by starting from the upper left and reading down and 188 around the corner. In this case, we get the result that 286×734=209,924. 189 190

Let’s compare this with a modern method of multiplication that is taught in 191 U.S schools (but not everywhere). With this method, each digit in the first 192 number is multiplied with each digit in the second. However, carrying 193 must take place for each multiplication where more than 10 is produced. 194 Any carrying leftovers must be added to the next product. Thus, 195 multiplication, carrying, and addition are all intermingled during the 196 multiplication process. Finally, everything can be added up, assuming that 197

zeros have been put into the proper locations so that all the place values line up correctly. 198 199 200 201 202 203 204 205 206 207 Example 1 208

Use the gelosia method to calculate 5108×327 209 Solution: 210

For this product we need a 4 by 3 grid. Multiplying first and then adding 211 gives: 212

213

214 215

So our result is 1,670,316. ♦ 216

2 8 6 × 7 3 4 1 1 4 4 8 5 8 0 2 0 0 2 0 0 2 0 9 9 2 4

Think About It

How are the two systems of multiplication alike? How are they different? What are the advantages and disadvantages of each? If

you showed each to someone unfamiliar with either method, which do you think would be easier to figure out?


217 Check Point A 218

Multiply 726×4138 219 220

221 222

Solution: 223 See endnotes to check your answer.1 224

225 Example 2 226

Use the gelosia method to compute 2.35×46.7. 227 Solution: 228

This method works perfectly fine when we have numbers with fractional 229 parts. We simply multiply as usual, ignoring the decimal point for the time 230 being. 231

232

233 234

We get a result of 109,745, which is obviously too 235 large for 2.35×46.7. Since we have three decimal 236 places to compensate for, we simply move the decimal 237 point three places to the left. We get 109.745,which is 238 correct. ♦ 239

240 This method found it’s way into 15th century Europe (via Islamic mathematicians and their 241 arithmetic books) when Luca Pacioli (1445−1517) included it in his popular book on arithmetic. 242 His comment on the origin of the method’s name is interesting not only because it tells us where 243 the word may come from but also because of the insight it gives into a part of the culture of Italy 244 at the time. 245

Think About It

How would you do 355 ÷ 4 with

this method?

Think About It

Why does moving the decimal point three places “fix” everything in this

example?


By gelosia we understand the grating which is the custom to place at the windows of 246 houses where ladies and nuns reside, so that they cannot be early seen. Many such 247 abound in the noble city of Venice.2 248

249 The method did not survive for long after the fifteenth 250 century. When printing presses were invented it is likely 251 that the method, with all of its grids and lines, was too 252 demanding on the first printing presses, so it gave way to 253 other algorithms that were more easily typeset. 254 255

Think About It

Our word jealousy comes from the word gelosia. Can you think of a reason why these two words would be

connected?


PART 2: The Solution of the Cubic and Quartic Equations 256 257 Historical and Mathematical Background 258 259 For many hundreds of years, mathematicians had methods of solving quadratic equations. 260 Eventually, they developed techniques that are equivalent to the modern quadratic formula. 261 Today, we know that if we have an equation of the form 02 =++ cbxax , we can find the 262 solutions of this equation algebraically, if they exist, with the quadratic formula: 263 264

aacbbx

242 −±−

= 265

266 By “finding the solution algebraically,” we mean that we can find numbers that satisfy the 267 equation by using the only basic operations of addition, subtraction, multiplication, division, 268 powers and roots. Various mathematicians had some very creative solutions to quadratics that 269 were not algebraic solutions, but they were valid nonetheless. However, there was something 270 very alluring about finding the equivalent of a formula for these kinds of equations. This allure 271 led many mathematicians to look for solutions to other kinds of polynomial equations. In 272 particular, the third and fourth-degree polynomial equations and their solutions were pursued 273 long and hard by mathematicians. 274 275 The algebraic solution of the cubic and quartic equations is one of the great highlights in all of 276 the history of mathematics. A cubic equation is one of the form: 277 278

3 2 0ax bx cx d+ + + = 279 280

The highest power of x is three. A quartic equation has four as the highest power of x and has 281 the form: 282

4 3 2 0ax bx cx dx e+ + + + = 283 284 By 1500, algebraic solutions to third and fourth degree equations had not been found, however, 285 despite valiant attempts to do so. In the 12th century, al−Khayyami had a method of solving an 286 equation of the form dcxx =+3 , but his method rested on seeing the equation as an equation 287 between solids. His solution was very geometric and today feels very “Greek” in nature. Al− 288 Khayyami actually examined 14 different kinds of cubic equations, described the physical 289 objects needed to solve each one, and then proved that each solution was correct.3 But this 290 approach was very different than the kind that was pursued later in history. Mathematicians 291 were not satisfied with a geometric approach; they wanted one that was more algebraic in nature. 292 293


Cardano and His Gang of Italian Algebraists 294 295 In 1494, Luca Pacioli, an Italian mathematician wrote a book called Summa de Arithmetica. In 296 this work, Pacioli discussed the solution of linear and quadratic equations. In this book, he 297 started using the word co, short for cosa (thing) to represent an unknown quantity. This was an 298 early version of our symbolic algebra system where a single letter, typically x, stands for an 299 unknown quantity. Pacioli also discussed the challenges of solving the cubic equation with an 300 algebraic approach and decided that it was probably NOT possible to do. In the next century, 301 however, many of his Italian counterparts did not share his opinion and attacked this problem 302 with great vigor. 303 304 At the University of Bologna, Scipione del Ferro (1465-1526) ignored Pacioli’s opinion and 305 discovered a formula for the solution to a cubic of the form: 306 307

3x mx n+ = 308 309

Pacioli did not publish his result. Instead, he kept it secret! To understand why he did this, it is 310 worthwhile to understand the Renaissance university of the time. This was a time when the 311 modern implementation of tenure did not exist. Jobs at universities were not secure like they are 312 today under tenure. To keep a post, you had to have not only political influence and the ability 313 to “shmooze” the appropriate people, but you also had to have the intellectual force to withstand 314 public challenges to your post. 315 316 These public challenges could occur at any time. In these scholarly battles, the current holder of 317 an academic post and his challenger would meet in public and match wits against each other. If 318 the challenger won, he would bring public humiliation to his counterpart, who would often have 319 to resign his post to his challenger. This was not a positive development in one’s career! 320 321 A new discovery like del Ferro had found was something to save in case of a challenge. If an 322 opponent appeared on the scene and had a list of problems to pose, del Ferro felt confident 323 enough that he could get at least some of them correct. On the other hand, if he was reasonably 324 sure that he was the only one with a solution to a cubic, he could counter his challenger’s list of 325 problems with a smattering of cubic equations to solve, thereby 326 giving him a decent chance of surviving the challenge. 327 328 As it turned out, del Ferro never needed to use his secret weapon. It is 329 reported that just before his death he passed on his solution to one of 330 his students, Antonio Fior. Unfortunately, Fior was not as prudent as 331 his master. Upon hearing that another mathematician, Niccolo 332 Fontana4, had boasted that he had found a solution to another form of 333 the cubic equation ( 3 2 )x mx n+ = , Fior immediately challenged 334 Fontana to a public contest. (Fontana was and is currently also 335 known as Tartaglia – the stammerer – due to the fact that he was 336 unable to speak clearly. Fontana had suffered a sword wound to his 337 face in 1512 when a French soldier attacked his hometown.) 338 339 Tartaglia


Tartaglia sent Fior a list of 30 varied problems. Fior, on the other hand, sent Tartaglia 30 cubic 340 equations, all of the form 3x mx n+ = . As the story goes, Tartaglia worked furiously on these 341 problems, presumably with his previous knowledge of how to solve equations of the form 342

3 2x mx n+ = available to help him. On February 13, 1535, Tartaglia cracked the solution and 343 was able to solve all 30 of Fior’s problems. Fior, of course, could not solve all of Tartaglia’s 344 problems as they were chosen more carefully. Gracefully, Tartaglia relieved Fior’s obligation as 345 the loser to shower Tartaglia with 30 banquets. However, the damage was done. Fior is no longer 346 remembered except for the foolishness he displayed during this challenge. 347 348 As word spread that Tartaglia had solved the solution to the cubic, it 349 eventually reached the ears of Gerolamo Cardano5 (1501-1576), one 350 of the most interesting characters in the entire history of mathematics. 351 Cardano was a doctor by trade, at one point serving the Pope, but 352 Cardano also worked extensively on mathematics. Cardano’s own 353 autobiography, De Vita Propria Liber (The Book of My Life), gives 354 us some idea of this man’s personality. He was a man who was 355 consumed with superstition and tragedy. There is much that could be 356 written about him, but readers are encouraged to do some basic 357 research and reading to get more information. (See Homework 358 Problem (85) for that opportunity.) 359 360 Cardano wrote to Tartaglia and asked him for the solution to the cubic 361 equation. Tartaglia, of course, initially refused. Why would he give away such a valuable secret? 362 But Cardano did not give up. He continued to write Tartaglia until he wore him down. Finally, 363 on March 25, 1539, Tartaglia revealed the method to Cardano in the form of a coded cipher after 364 Cardano agreed to take the following oath:6 365 366

I swear to you by the Sacred Gospel, and on my faith as a gentleman, 367 not only never to publish your discoveries, if you tell them to me, but I 368 also promise and pledge my faith as a true Christian to put them down 369 in cipher so that after my death no one shall be able to understand 370 them. 371 372

Along with a student, Lodovico Ferrari (1522-1565), Cardano unraveled the secret of the cubic 373 equation in the form nmxx =+3 , making significant progress. Not only did they master the 374 techniques given to them by Tartaglia, but they also extended his work to apply to cubic 375 equations of any form (a huge accomplishment) and used Tartaglia’s techniques to solve 376 polynomial equations of degree four! Cardano and Ferrari were bound by Cardano’s oath, 377 however. Even though they had pushed Tartaglia’s work far beyond where Tartaglia had ever 378 taken it, all of their progress was based on the oath-protected secrets of Tartaglia. Cardano, 379 though, was anxious to publish his results. He did not need to protect an academic post since he 380 was a doctor by trade. 381 382 Looking for a way out of this predicament, Cardano and Ferrari traveled to Bologna, where our 383 story began. There, they studied the private papers of del Ferro and saw the exact solution of 384 Tartaglia written in del Ferro’s own handwriting, but well before Tartaglia had discovered them 385

Cardano


independently for himself. As far as Cardano was concerned, he was no longer bound by his 386 oath. He would publish del Ferro’s findings, not Tartaglia’s, even though they were the same. 387 And that he did. In 1545, he published Ars Magna (Great Art), which is considered one of the 388 great masterpieces in mathematical history. (It’s still in print!) He prefaced his work with the 389 following attribution: 390 391

Scipio Ferro of Bologna well-nigh thirty years ago discovered this rule 392 and handed it on to Antonio Maria Fior of Venice, whose contest with 393 Niccolo Tartaglia of Brescia gave Niccolo occasion to discover it. He 394 gave it to me in response to my entreaties, though withholding the 395 demonstration. Armed with this assistance, I sought out its 396 demonstration in [various] forms. This was very difficult.7 397

398 Tartaglia was furious, of course. He accused Cardano of deceit, sending nasty letters to Cardano 399 with the charges clearly spelled out. Cardano basically ignored them. His student, however, was 400 more easily drawn into the debate. In 1548, Tartaglia and Ferrari squared off in a public, 401 mathematical challenge in Milan, Ferrari’s stomping grounds. Tartaglia lost, blaming his 402 performance on the “rowdiness and partisanship of the crowd.”8 (A rowdy crowd at a math 403 contest?) Tartaglia went back home, defeated once again, and Ferrari was proclaimed the winner. 404 405 Ars Magna and the Solution of the Cubic 406 407 In Ars Magna (1545), Cardano detailed del Ferro’s 408 and Tartaglia’s technique for finding the solution of a 409 cubic equation in the form: 410

411 nmxx =+3 412

413 This is called a depressed cubic because the 2x term 414 is missing. Cardano’s solution was given entirely in 415 words. There were no modern algebraic symbols yet 416 present to help him express his results. In modern 417 notation, Cardano’s verbal solution of this kind of 418 cubic takes on the following modern form: 419 420 421

422 423 424

425 426

427 428

429 430 431

Cardano’s Formula

3 3

2 2n nx R R= + + −

where 2 3

4 27n mR = +


You can guess why this equation is not introduced into basic algebra courses. There is relatively 432 little that is “basic” about it. Prior to the advent of handheld calculators, using this formula 433 would prove to be almost impossible in most cases. 434 435 The demonstration of this formula is not an easy task to undertake.9 However, it is based on 436 basic algebraic principles and so if you are patient and diligent enough, you can follow how it 437 was derived. In this text, we will skip an explanation of where the equation comes from and 438 focus our efforts on learning how to use it. 439 440 (Note: In order to use this formula, you need to really know how to use your scientific or 441 graphing calculator. To take cube roots on your calculator you can check to see if your model has 442 a 3 x button. If it does not, you can take a cube root by raising a number to the power of 1/3. For 443 example, 28 3/1 = . On a calculator, the power button looks usually like xy , yx , or the ^ symbol. 444 To do 81/3 would require a key sequence like: 8^(1/3), or 8 yx (1/3). Play with it on your model 445 until get 2 for 81/3 before you proceed.) 446 447 Before we do any specific examples, it might be helpful to propose some steps that can be 448 followed to use this formula correctly: 449 450 Steps in Using Cardano’s Formula 451

Step 1: Make sure that the equation given is in the form nmxx =+3 . 452 Step 2: Identify the proper values of m and n. 453 Step 3: Calculate the value of R and also R . Simplify if possible. 454 Step 4: Plug in the value for R into Cardano’s equation and carefully compute the 455

final value. 456 457 We will start with the equation that Cardano used to illustrate his method. He stated his problem 458 in words (since he did not have variables to work with) in a form that might look like this: 459

460 Cube plus six times a number is twenty. What is the number? 461

462 A translation of this into modern algebraic notation would be the following: 463 464

2063 =+ xx . 465 466 467 468

Cube plus six times a number is twenty. What is the number? 469 470


Example 3 471 Use Cardano’s formula to solve 2063 =+ xx . 472

Solution: 473 Step 1: Make sure that the equation given is in the form nmxx =+3 . This 474

equation is already in this form so we can proceed. 475 476 Step 2: Identify the proper values of m and n. 477 The value of m is 6 and the value of n is 20. 478 479 Step 3: Calculate the value of R and also R . Simplify if possible. 480

2 3

2 3

4 2720 64 27

400 2164 27

100 8108

n mR = +

= +

= +

= +=

481

482 This means that 108R = 483

484 Step 4: Plug in the value for R into Cardano’s equation and carefully 485

compute the final value of x: 486 487

3 3

3 3

3 3

2 220 20108 1082 2

10 108 10 108

n nx R R= + + −

= + + −

= + + −

488

489 This is what we could call the exact solution. If we want a decimal 490

approximation to this, we can use a calculator to do this: 491 492

3 3

3 3

3 3

10 108 10 108

10 10.3923 10 10.3923

20.3923 0.39232.7321 0.73202.0001

x = + + −

≈ + + −

≈ + −≈ −≈

493

494

It’s amazing that these two are actually equal.


We get an estimate of 2.0001. This is pretty close to 2, so we’ll 495 round it off to 2 exactly and see how close we are. If we check this 496 in the original equation, we get: 497

498

20128

2626 33

=+=

×+=+ xx 499

500 We do get the correct value. ♦ 501 502 503

504 505 506 507 508 509 510 511 512 513 514 515 Note that this equation gives one solution for this 516 equation. We can look at the graphs of xxy 63 += and 517

20=y , to see where they intersect since the 518 intersection point(s) represent solutions of the original 519 equation. The graph here shows that there is indeed 520 one solution. You can see that when x = 2, the curve of 521

xxy 63 += crosses the line y = 20, just as we expect it 522 to. Obviously, Cardano did not have the advantage of 523 seeing such a graph…the x−y coordinate system was 524 not yet invented, but it does give us some insight into 525 what is going on here. 526 527 Example 4 528

Use the cubic formula to solve 1033 =+ xx 529 Solution: 530

Step 1: Make sure that the equation given is in the form nmxx =+3 . 531 This equation is already in this form so we can proceed. 532

533 Step 2: Identify the proper values of m and n. 534 The value of m is 3 and the value of n is 10. 535 536

-2 0 2 4 6

-10

0

10

20

30

Important Note

In order to be as accurate as possible with these calculations, it is best to keep as many decimal places as possible. If you know how, it is best to try to do the entire sequence of

calculations without clearing our your calculator or using the clear button. (By not using the clear button, my graphing calculator gives me the exact value of 2!) You

should experiment with your calculator and try to get an exact value without having to use the clear button. If you cannot get it to work, use four decimal places throughout

your computations.

(2,20)


Step 3: Calculate the value of R and also R . Simplify if possible. 537 538

2 3

2 3

4 2710 3

4 27100 27

4 2725 126

n mR = +

= +

= +

= +=

539

540 This means that 26R = 541

542 Step 4: Plug in the value for R into Cardano’s equation and carefully 543

compute the final value 544 545

3 3

3 3

3 3

2 210 1026 262 2

5 26 5 26

n nx R R= + + −

= + + −

= + + −

546

547 This is we call the exact solution; the decimal approximation 548 is: 549

550

6989.146263765.161522.2

09902.009902.10

09902.5509902.55

265265

33

33

33

≈−≈

−≈

+−−+≈

+−−+=x

551

552 A check of this shows a close match, although the decimal 553 rounding causes a little error. ♦ 554

555 Check Point B 556

Use the cubic formula to solve 3 10 57x x+ = 557 Solution: 558

See endnote for an answer.10 559 560


Check Point C 561 Use the cubic formula to solve 3 5 15x x+ = 562

Solution: 563 Seen endnote for an answer.11 564

565 566 Solving the General Cubic Equation 567 568 Cardano next tackled the non−depressed equation. To do this, he came up with a clever way of 569 temporarily converting a non−depressed equation into one that is depressed. He then used his 570 formula on the depressed equation and then adjusted his result to reflect the fact that he had 571 changed the original equation. 572 573 The general method is as follows: 574

575 Solving the Non-Depressed Cubic Equation: 576 Given the equation 023 =+++ dcxbxax 577

Step1 Let abyx3

−= . 578

Step2 Substitute abyx3

−= into the original equation to get an equation 579

in y. The result should be a depressed cubic. 580 581

Step3 Use Cardano’s formula to solve the equation for y. 582

Step4 Find the original, desired values of x using abyx3

−= . 583

584 585 Example 5 586

Solve 3 22 30 162 350 0x x x− + − = 587 Solution: 588

Step 1: Let 5)2(3

303

+=−

−=−= yyabyx . Remember this for later: 5x y= + 589

Step 2: Substitute x = y + 5 into the original equation to get a depressed cubic. 590 591 592 593

3 2 3 2

3 2 2

3 2 2

3

2 30 162 350 2( ) 30( ) 162( ) 3502( 15 75 125) 30( 10 25) 162 810 3502 30 150 250 30 300 750 162 810 3502 12 4

5 5 5

0

x x xy y y y y y

y y y y y y

y

y

y y

y

− + − = − + −

= + + + − + + + + −

= + + + − − − + + −

= +

+

−

+ +

594

595

x = y + 5


Recall that this is equal to 0, so we can divide by 2 to get: 596 597

3

3

3

2 12 40 06 20 06 20

y yy yy y

+ − =

+ − =

+ =

598

599 Step 3: This is a depressed cubic and we can now use Cardano’s formula on 600

it. But note that this is essentially the equation Cardano used to 601 illustrate the solution to the cubic. We saw in Example 3 that 602

2063 =+ xx has a solution of 2, therefore 2063 =+ yy has a 603 solution of y=2. 604

605 Step 4: Since x = y + 5, and y = 2, we can see that x = 2 + 5 = 7. Thus, x = 7 606

is the solution to the original equation 3 22 30 162 350 0x x x− + − = . 607 608

Check: 609 610

( ) ( ) ( )3 32 7 30 7 162 7 350 686 1470 1134 3500

− + − = − + −

= 611

612 This verifies the solution is correct. ♦ 613

614 Using this process may seem cumbersome to some, but we should remember that Cardano did 615 not have variables or equations to work with. Cardano had to mainly use words to solve these 616 equations. Hence, the process that we have is actually a lot easier to use than Cardano’s. 617 618 Example 6 619

Solve 018156 23 =−+− xxx 620 Solution: 621

Step1: Let ( 6) 23

x y y−= − = + 622

623 Step2: Substitute x in to the equation: 624

625

43043

018)2(15)2(6)2(018156

3

3

23

23

=+

=−+

=−+++−+

=−+−

yyyy

yyyxxx

626

627 This last equation is in the form we need to use Cardano’s formula. 628 629


Step3: We have m = 3, n = 4. 630 We first find the values of R and R . 631

632

2 3

2 3

4 274 34 27

4 15

n mR = +

= +

= +=

633

634 Thus, 5R = . 635 636 Now find the value of y using Cardano’s formula. 637

638

3 3

3 3

3 3

2 24 45 52 2

2 5 2 5

n ny R R= + + −

= + + −

= + + −

639

640 Using a calculator, we can get an estimate for this number: 641

642

3 32 5 2 5 1.6180 ( .6180)

1+ + − ≈ + −

= 643

Therefore, y = 1. This means that y = 1 is a solution to 3 3 4y y+ = . 644 Keep in mind that our original equation was in terms of x and that we 645 used the substitution 2x y= + . We now use this substitution again to 646 find the value of x that we are after. 647 648 Step4: y = 1, so x = y + 2 = 1+ 2 = 3. Hence, x = 3. 649 650 Check: 651 652 We let x = 3 and substitute in the original cubic equation. 653

654 ( ) ( )

07272

184554271831536318156 2323

=−=

−+−=−+−=−+− xxx

655

656 This solution checks and we’re done.♦ 657


658 Check Point D 659

Use the cubic formula to solve 3 212 4 445 0x x x+ + − = 660 Solution: 661

See the endnotes for a solution.12 662 663 664 The Solution of the Quartic Equation and Beyond 665 666 With the cubic equation conquered, Cardano and his student, Ferrari, moved on to the quartic 667 equation of the form 4 3 2 0ax bx cx dx e+ + + + = . Could a formula be found for that as well? The 668 answer was yes, as long as the equation could be reduced to a depressed cubic, in which case 669 they could use their previous discovery. It seems as though the cubic formula was more useful 670 than they originally thought. We won’t explore the solution of the fourth degree equation in 671 detail here due to its complexity. However, if you’re interested in seeing a modern 672 representation to the solution of the quartic, you can read Appendix A at the end of the chapter. 673 674 Fifteen years after Cardano and Ferrari did their work, Rafael Bombelli (1526−1572) wrote a 675 more systematic text to help students master these techniques. His book, Algebra, “marks the 676 high point of the Italian algebra of the Renaissance.”13 In this text, Bombelli introduced 677 equations and techniques that led to the establishment of what we call imaginary or complex 678 numbers and gave rules for working with them. Complex numbers include numbers of the form 679

bia + where i = 1− . That is, the square root of a negative number is allowed and is considered 680 a valid number. (These are not part of the real number system. In fact, the real numbers are 681 actually a subset of the complex numbers.) Complex numbers play important roles in higher 682 mathematics, physics, and engineering. 683 684 We make one last note about solving equations: It was only natural for Cardano and Ferrari to 685 pursue the quintic (fifth degree) equation and its solution. A quintic equation has the following 686 form: 687

5 4 3 2 0ax bx cx dx ex f+ + + + + = 688 689

But they were thwarted. They could not find an equation to solve a quintic 690 because there is none. It is impossible to express the solution of a fifth−degree 691 equation with a formula using roots, powers, and the four basic mathematical 692 operations. This was first proved by Niels H. Abel (1802−1829). Later 693 Evariste Galois (1811−1832) extended his work and gave general conditions 694 for when equations are solvable. Galois has a whole field of mathematics 695 named after his work…Galois Theory. Unfortunately, both Abel (top) and 696 Galois (bottom) died at very early ages. The story of Galois’ life and death is 697 fit for a television movie of the week. 698 699 As the 1500’s and 1600’s passed, mathematics was rediscovered in Europe 700 again. It stands on the shoulders of many civilizations that preceded them. By 701 the early 1600’s, Europe was ready for new techniques that would help them 702 tackle modern, complex calculations. 703


PART 3: Modern Calculations 704 705 The power and extent of modern calculations rest on at least three important developments.14 706 The introduction of the Hindu−Arabic numbers, the use and acceptance of numbers with 707 decimal/fractional parts, and the invention of logarithms all played important roles in how 708 mathematicians and scientists could do computations. The introduction of decimal fractions and 709 logarithms both took place in the early 1600’s and each of them came about because of efforts to 710 take very laborious computational techniques and simplify them. (Prior to this, very complicated 711 methods from trigonometry had been widely used instead.) The need for this kind of change was 712 motivated by the needs of research in astronomy, navigation, and commerce, to name a few. 713 714 Decimal Fractions 715 716 Simon Stevin15 (1548−1620) was one of the leading advocates of 717 decimal fractions. Just to clarify what we mean by “decimal 718 fractions,” we are talking about numbers like 34.657. These are 719 base−ten numbers where the whole part and fractional part of a 720 number are taken together and computations are done on them 721 without splitting up their individual parts. Prior to this time, the 722 whole and fractional part were often split apart and computed 723 separately. Not only that, but fractions were often expressed in base 724 60 rather than in base 10. 725 726 727 728 729 730 731 Stevin worked in commercial, military, and administrative environments, and eventually taught 732 math at the Leiden School of Engineering. He was one of the first people to fully embrace the 733 emerging Copernican theory. He gains a role in this discussion because he published a pamphlet 734 called De Thiende (The Art of Tenths) where he introduced decimal fractions. His publication 735 came at a time when others had started taking advantage of decimal numbers, and so it was 736 received with open arms. It is interesting to see the difference between his notation and ours. 737 He used the symbols , , , , … to represent descending powers of 10. 738 739

Corresponds to the whole part of a number (ones place and higher, basically) 740 Corresponds to the tenths place; (1/10)1 is tenths. 741 Corresponds to the hundredths place; (1/10)2 is hundredths 742 Corresponds to the thousandths place; (1/10)3 is thousandths, etc…. 743

744

Think About It Why Base 60?


However, instead of using “tenth,” “hundredth”, or “thousandth,” Stevin uses “prime,” “second,” 745 and “third,” respectively. If he wanted to express the number 34.657 it would look similar to this: 746

34 6 5 7 747 Here’s a breakdown to clarify his notation: 748

34 6 5 7 749 750

751 34 whole 6 in the tenths (prime) place 5 in the hundredths (second) place 7 in the thousandths (third) place 752 753 754 Example 7 755

Write 287.93 in Stevin notation 756 Solution: 757

287 9 3 758 759 Check Point E 760

Write 3,456.28547 in Stevin notation 761 Solution: 762

See endnotes for solution.16 763 764 If Steven wanted to do arithmetic with decimal fractions, an addition 765 problem would look like the picture here, which is actually a piece of 766 Stevin’s publication on this subject. 767 768 Around this time, support for these numbers grew quickly17, despite their 769 clunky implementation. In 1592, Giovanni Antonio Magini, a mapmaker, 770 introduced the decimal point to separate the whole parts from the 771 fractional part of numbers. Despite its ongoing evolution and staunch 772 support for the system from people like Stevin, it would be another 200 773 years before the system was adopted for use in currency, weights, measures, etc. 774 It was not until the French Revolution that the “metric” system was adopted for 775 such uses. 776

777 Another important name that paved the road for decimal 778 fractions and logarithms to emerge was François Viète18 779 (1540−1603). Viète, convinced that a geocentric model 780 was more reliable than the emerging Copernican theory, 781 made great efforts to publish trigonometry calculations 782 that would help in the study of astronomy. For the most part, scientists 783 before him were used to using sexigesimal (base-60) fractions that were 784 a product of the Babylonian number system and their own work in 785 astronomy. Viète, along with more and more of his contemporaries, was 786 advocating a new system of decimal (base 10) fractions. 787

788 One way in which Viète and others tried to simplify multiplication and division was by reducing 789 them to addition and subtraction. The obvious reason for this was that adding and subtracting 790 numbers is generally much easier than multiplying or dividing them. This process of reduction is 791

Think About It

What numbers are being added here and what is the

result?


called prosthaphaersis.19 In his publication, Canon, Viète presents several equations from 792 trigonometry that accomplish this. For example, he gives the following two rules: 793 794

)sin()sin(sincos2)sin()sin(sinsin2

BABABABABABA

−−+=×−++=×

795

796 The “sin” and “cos” symbols are trigonometry functions that are actually abbreviations for sine 797 and cosine, respectively. The phrase )cos( BA + does not mean cos “times” (A+B). Instead, it 798 means the cosine of (A+B). It’s sort of like a square root. The symbol has no meaning by 799

itself…we have to take the square root of some particular number. BA + does have meaning 800 (usually). 801 802 While we will not study these trigonometry functions in this course, what we will point out is 803 that each of these two equations reduces multiplication into addition (and/or subtraction). The 804 first equation takes two sines that are multiplied together and reduces the value to two sines that 805 are added together. With tables to give the values of these cosines and sines, computations could 806 (and did) go much more smoothly. We will see that this idea of reducing multiplication and 807 division to addition and subtraction is an important one in the next part of this chapter. 808


PART 4: Napier and The Emergence of Logarithms 809 810 The mere mention of the word “logarithm” can cause many algebra students’ eyes to gloss over. 811 Most do not understand these tools when they first see them and rarely remember them past the 812 final exam. I’m convinced that a big part of the reason that the rules and uses of logarithms are 813 so hard for students to grasp is that they are usually presented completely devoid of their 814 historical context. Furthermore, with the advent of technology, their general use has been 815 reduced significantly. In this section, we’ll try to establish that context and hopefully come to see 816 logarithms as an important and powerful development in the history of math (even if they have 817 been replaced by the modern calculator). 818 819 Keep in mind that we are considering a time in history when calculators 820 were not present and calculations all had to be done “by hand.” As work in 821 astronomy and other fields progressed, the need for an efficient method of 822 calculation had emerged. As we stated before, in some circles this was 823 done with prosthaphaersis, where multiplication and division were 824 reduced to addition and subtraction. John Napier20 (1550−1617) spent 825 about twenty years of his life developing a new and easy-to-use tool that 826 would use this approach to make computations easier to do. He gave it the 827 name logarithm. 828 829 Napier was an ardent Scottish Presbyterian who was strongly opposed to Catholicism. In his 830 book A Plaine Discovery of the Whole Revelation of Saint John: Set Downe in Two Treatises, he 831 identifies the Pope as the antichrist, urges King James to rid his house of “papists,” and predicts 832 the end of the world will take place between 1688 and 1700. He was also a resourceful 833 landowner, devising a hydraulic screw similar to that of Archimedes to control water levels in 834 coal pits. On the mathematical front, he used numerology to look for hidden prophecies in the 835 Bible and he looked for a way to simplify computations in trigonometry. To do this, he used a 836 well−known trigonometric equation of the time: 837

838 )cos()cos(sinsin2 BABABA +−−=× 839

840 Although slightly different from the equations given 841 previously, we can see that the multiplication on the 842 left is reduced to addition and subtraction on the right. 843 Napier used this identity to build a table of logarithms 844 and “antilogarithms” that could be used to simplify 845 calculations. In 1614, he announced and published his 846 tables in Mirifici Logarithmorum Canonis Descriptio 847 (A Description of the Marvelous Laws of Logarithms). 848 Later, he published information on how he developed 849 logarithms and the theory behind them in Mirifici 850 logarithmorum canonis constructio (A Construction of 851 the Marvelous Laws of Logarithms.) The word logos 852 means ratio in this context and arithmos means 853 number; Napier merged them to form the word 854


logarithm (ratio of number). The reason he chose this particular word has to do with how he 855 came up with the logarithm concept and later developed it. 856 857 It is important to point out that Napier’s logarithms are different than our own. It wasn’t until 858 after his tools had become widespread and had been refined and developed by others that they 859 took the form that they are in today. To Napier, logarithms were tied to physical motion and the 860 mathematics related to motion. Napier defines his logarithms as follows:21 Take a line segment 861 AB and an infinite ray DE, as shown below: 862

863 864 865 866 867 868 869 870

871 872 873 874 Let C be a point that starts at point A and moves along AB. Let F be a point that starts at point D 875 and moves along ray DE. Points C and F both start moving at the same speed. From that point 876 on, their speeds change according to the following rules: 877 878

Point C always moves with speed equal to the distance CB = y. 879 Point F always moves at the same speed as when it started, 880 where x is the total distance F has moved from point D. 881

882 According to Napier, x is the logarithm of y. 883

884 This definition looks very little like the modern definition of logarithms because the modern idea 885 of log didn’t come along until the time of Euler (1707-1783). We’ll explore the modern 886 definition more in a moment, but what we should try to keep in mind is that both versions of the 887 tool are trying to reduce multiplication and division to addition and subtraction. Using this (or 888 the modern) definition of logarithm, it is possible to build tables that will make these reductions 889 for us. Napier spent 20 years of his life painstakingly developing this theory and building his 890 tables to an incredible degree of accuracy. He made very few errors, which is amazing 891 considering the number and complexity of calculations that he undertook. 892 893 Astronomers and mathematicians alike were ecstatic about this new invention since it really did 894 do what Napier had intended. One of the most famous quotes about logs comes from 895 Pierre−Simon who stated that “[logarithms], by shortening labors [on computations], doubled the 896 life of astronomers.” The acceptance of Napier’s new tool was widespread and rapid. Rarely in 897 mathematics are inventions so quickly and universally adopted. 898 899

A B

D E

C

F

y

x

Think About It

Is point C slowing down or speeding

up as it moves along AB? Why?


In 1616, Henry Briggs (1561−1631) worked with Napier to refine his invention so that they were 900 more convenient to use. Briggs is known for helping to construct tables where log 10 = 1, thus 901 creating what we today call the “common logarithm.” This is the log button you see on your 902 scientific or graphing calculator. Briggs published common log tables for values from 1 to 1000. 903 Later, others provided tables up to 100,000 to ten places of accuracy! These were in use all the 904 way until the early 1900’s when the tables were computed to twenty decimal places in Britain. 905 This was still before the age of the electronic calculator or computer! 906 907 Napier’s work had long−range implications for broader mathematics, navigation, astronomy, 908 banking, and number theory. In an indirect route, the logarithm eventually led down a path that 909 resulted in the final proof that the three great Greek construction problems could not be solved 910 by straightedge and compass alone.22 911 912 The Idea and Use of Logarithms 913 914 In this section we will explore basic ideas of how logarithms work and why they were useful in 915 the time of Napier. This will hopefully explain why they were so popular. Furthermore, it is my 916 hope that the “rules of logarithms” that may have perplexed you in the past will be placed into a 917 proper context and that you will see how and why they were useful. 918 919 Recall that Napier (and others after him) published tables of logarithms and antilogarithms that 920 were intended to make complex calculations easier. As we work our way through this section, try 921 to imagine yourself in a world where you cannot reach for a calculator to do any number 922 crunching for you. Furthermore, the modern algorithms for multiplication and division that you 923 learned as a child do not exist. With these restrictions, how would you do calculations like the 924 following? 925 926

(34,879.2945) (1,998,736,882.812)× 927 928

2.57104.2 12× 929

930 These are certainly not computations we want to undertake without our little button−filled slabs 931 of plastic. But Napier and others in his time had no choice. Logarithms gave them hope of doing 932 these calculations more efficiently. 933 934 The Idea Behind Logs 935 936 Here’s the main idea behind logarithms and how Napier intended them to be used: We start with 937 a problem that involves complex arithmetic computations (like those given above). After 938 applying logarithm rules we get a new statement that we can look up on the logarithm tables. We 939 do some addition and subtraction operations on this statement (as needed) to get an intermediate 940 result. This intermediate result can be looked up in an antilog table to get us back to a result that 941 compensates for the fact that we reduced everything to addition and subtraction. Think of the 942 antilog table as the “evil twin” of the logarithm. What the logarithm does, its evil twin (the 943


antilog) comes along and undoes. So while the log will reduce to addition and subtraction, the 944 antilog will covert back to multiplication and division. Here’s a diagram of what we mean: 945 946 947 948 949 950 951 952 953 954 955 956 957 Rule of Logs…One Step Closer to the Tables 958 959 Well, we’ve put it off long enough. We can’t go any further until we talk about the rules of logs. 960 We will be working with modern versions of these rules that were unknown to Napier. But the 961 spirit of the task at hand remains the same. 962 963 The modern definition of the logarithm is defined in a way so that exponents are undone. For 964 example, if we have the equation 252 =x , we know that we can take the square root of both 965 sides, as square roots undo powers of 2. If the equation looks like 252 =x , taking a square root 966 no longer works since the exponent (x) is not a 2. In fact, it’s a variable! In order to undo this 967 exponent, we need a tool that will specifically deal with this kind of equation. Leonhard Euler 968 (1707−1783) is responsible for the definition of log that we will use: 969 970

log pb n p b n⇔= = 971

972 The ⇔ symbol here is used to imply interchangeability. Both sides of the double arrow lead to 973 each other and allow you to go back and forth between the two expressions. We often say that 974 the “log base b of a number n is the power to which b must be raised to produce n.” This 975 statement tells us what taking the log of a number (n) means. The number b is the base. The 976 number p is the result. You should observe that while the right side of the double arrow has a 977 power present, the left side does not. The logarithm has “undone” the power. 978 979 Some simple review examples are in order. 980 981 Example 8 982

Write 823 = as a log statement. 983 Solution 984

2 = b, the base; 3 = p, the power; 8 = n, the number that is produced. 985 So, 2log 8 3= 986

We say that the “log base 2 of 8 is 3.” That is, the power to which 2 must be 987 raised to get 8 is 3. ♦ 988

Ugly computation is converted to a

statement using logs

Log rules/tables are used so

addition and subtraction yield an intermediate

result

Antilog tables are used to convert the intermediate result to base 10 numbers…the

final result

Via log

table

Via anti-log

table


989 Example 9 990

Write 252 =x as a log statement. 991 Solution 992

2 = b, the base; x is the power; 25 = n, the number. 993 So we have 2log 25 x= . ♦ 994 995

Check Point F 996 Write 4 16y = as a log statement. What is y? 997

Solution 998 See the endnotes to check your answer.23 999

1000 Example 10 1001

Write x=49log7 as a statement with exponents. 1002 Solution: 7 = b, the base; 49 = n, the number produced; x is the power required. So 1003

we get 7 49x = . We can actually find the value of x here, since we know that 1004 72 = 49. Thus x = 2. ♦ 1005

1006 Check Point G 1007

Write wzy =log as a statement with exponents 1008 Solution 1009

See the endnotes to check your answer.24 1010 1011 So, what does this have to do with making multiplication and division easier? Good question. 1012 Unless otherwise specified, we will assume from now on that our base is 10. Hence we will be 1013 working with the common logarithm that Briggs developed with Napier. If you don’t see a 1014 base on a log, you can assume the base is 10. (By the way, that’s what your calculator log button 1015 assumes.) Let’s take two numbers and multiply them. We’ll take two special numbers that will 1016 help us demonstrate the first log rule, which reduces multiplication to addition. 1017 1018

Let xA 10= and let yB 10= 1019 1020 A and B are two numbers, which depend on x and y, of course. Let’s see what happens when we 1021 multiply them: 1022


Operation Comment yxAB 1010 ×= Multiply our numbers

( )

1010

x y

x y

ABAB

+

+

=

=

Recall that by rules of exponents, when we multiply the same base with different exponents, we add the exponents and keep the base. For example 64242 xxxx ==⋅ + . The parentheses around x + y emphasize that we can consider this all one exponent.

10 (og )l AB x y= + We simply apply the modern definition of the log, where 10 is the base, (x+y) is the exponent, and AB is the number/result.

If we knew what x and y were, we could proceed. Although we don’t know what values they take on, we can use the modern log definition to find expressions for each.

10

10

log

lo

10

g10

x

y

A

B

A x

B y

= ⇔

=

=

=⇔

Apply the modern definition of the log to A and B.

Now comes the cool part…

1 10

10

0l log

ogl

ogA x

A BB y= +

= +

Substitute our values for x and y into the new equation.

1023 This gives the first famous rule of logs: 1024 1025

1026 1027 1028

We’ve omitted the base since this rule is true for all valid bases. No doubt you’ve seen this 1029 before, unaware of where it came from or why it even showed up in the first place. It’s often 1030 paraphrased as “The log of a product is the sum of the logs.” With a historical context in place, 1031 we can see its importance. The left side is the log of two numbers that are multiplied. On the 1032 right side, we have reduced it to two logs being added together! Aha. There it is again. We’ve 1033 reduced multiplication to addition and subtraction, just like Napier was seeking to do. 1034 1035 We can look at a simple example to see how this law works: 1036 1037 Example 11 1038

Write )4435log( × as the sum of logs. 1039 Solution 1040

44log34log)4435log( +=× 1041 1042 If we knew the value of log 34 and the value of log 44, we could add them to 1043 get the value of log(35×44). ♦ 1044 1045

1046

BAAB logloglog +=


Check Point H 1047 Write log(1298 3982)× as the sum of logs. 1048

Solution 1049 See the endnotes for solution.25 1050 1051

These examples so far don’t truly allow us to peer into the power of this rule. Here’s an example 1052 that will begin to do that. 1053 1054 Example 12 1055

What is z = 1654×3298? 1056 Solution 1057

Here we have a multiplication problem. It’s not hard. We can do it in a snap. 1058 But let’s just pretend that these two numbers are very hard to multiply. Let’s 1059 let x = 1654 and let y = 3298. 1060

1061 Operation Comments

32981654×=z This is what we want to compute, but we’re assuming it’s too hard or too time-consuming to do “by hand.”

)32981654log(log ×=z We take the log of both sides. In this step we are taking an “ugly computation and converting it to a statement about logs.” (See diagram above).

3298log1654loglog +=z Apply the first log rule to the right side. “The log of a product is the sum of the logs.”

7367.65182.32185.3log

=+=z Use the log tables to look up log1654 and

log3298. We then easily add up the results. What tables, you ask? You’ll meet them soon enough. We’ll finish this first.

7367.6log =z Here is our final result, in “logland.” z = 5453809.951 Use an antilog table to get z. Remember, the

antilog is the evil twin of the log, so it undoes the what the log does. In this case, it removes the log from the z and returns the value of z. (Again, the antilog tables are on their way.)

1062 When we check our answer, we are just slightly off, but we only worked with 1063 about four decimal places. Recall that the British had tables of twenty places 1064 in place in the early 1900’s. ♦ 1065

1066 1067


1068 1069 1070 1071 1072 1073 1074

1075 1076 1077 Hopefully you see from this example the process we described in the diagram above. We start by 1078 creating a statement about logs. We then use log rules and tables to create results that are easily 1079 added. Finally, we can use an antilog table to get us out of “logland” and to a final answer. 1080 Before we introduce the log and antilog tables, we should mention two more log rules that are 1081 important. The first of these is: 1082 1083

1084 1085 1086

This is the rule that reduces division to subtraction. It is related to the familiar rule of exponents 1087

that says nmn

m

xxx −= . That is, when we divide exponents with the same base, we subtract the 1088

exponents. This is why 62

8

xxx

= instead of x4. We don’t divide the exponents. The proof of this 1089

law is left as an exercise. 1090 1091 1092 Example 13 1093

Simplify 95.2log8

. 1094

Solution 1095 95.2log log95.2 log8

8 = −

. Hence, if we can compute log95.2 and log8, we 1096

can subtract them to get the value of the original log. ♦ 1097 1098

Example 14 1099

Simplify 85.3logx

. 1100

Solution: 1101 85.3log log85.3 log x

x = −

♦ 1102

1103

Ugly computation is converted to a

statement using logs

Log rules/tables are used so

addition and subtraction yield an intermediate

result

Antilog tables are used to convert the intermediate result to base 10 numbers…the

final result

Via log

table

Via anti-log

table

BABA logloglog −=


Check Point I 1104

Simplify log3.5z

. 1105

Solution: 1106 See endnotes for solution.26 1107

1108 The third log rule is also useful and is related to the familiar rule of exponents: ( ) mnnm xx = . For 1109

example, ( ) 155335 xxx == × . This rule states that when we raise an exponent to a power, we 1110 multiply the powers. The related log rule is: 1111 1112 1113

1114 1115

Many people remember this rule by observing that all it says is that “in a logarithm the power of 1116 the number can move to the front of the log.” 1117 1118 Example 15 1119

Simplify 3log x . 1120 Solution 1121

3log 3logx x= ♦ 1122 1123

1124 Example 16 1125

Simplify log x . 1126 Solution 1127

Recall that 1/ 2x x= 1128 1log log2

x x= ♦ 1129

1130 1131 Check Point J 1132

Simplify 23log x . 1133 Solution 1134

See endnotes for solution.27 1135 1136

ncn bc

b loglog =


Let’s prove this last log rule, ncn bc

b loglog = 1137 1138 Operation Comments Let nx blog= We start in this unusual place…but be patient

and you’ll see how things work out. nbx = Use the modern definition of log. We file this

away for use later. Consider log c

b n Now we look at the log we are seeking to simplify.

loglog ( )x cb b

cn b= Substitute xbn = from the step above. = ( )xc

b blog Multiply exponents according to rules for exponents.

= xc This one is a little subtle. Remember that the log base b of a number is the power that b must be raised to get that number. In this case our base is b and we want to raise that to some power to get our number, xcb . Well, of course, to get xcb we need to raise b to the power of xc, so xc is the log.

= cx logbc n=

So log logcb bn c n=

Here we just substitute the value of x from the very beginning to get our desired result.

1139 1140 The Log Tables 1141 1142 Most of you reading this have probably never seen a log table before. But before the relatively 1143 recent appearance of the handheld calculator and computer, log tables and their more 1144 sophisticated counterparts, slide rules, were the dominant tools for complex calculations. 1145 Using a log table, we can do sophisticated problems of multiplication and division, as well as 1146 powers and roots, without a calculator at our side. So for this section, you’ll want to put away 1147 your calculator and journey back in time. For the most part, we will limit ourselves to 1148 computations with numbers that have three or four decimal places. This will make using tables a 1149 little easier than they might be otherwise. Also, we will be using base-10 logarithms for all of 1150 our computations. You should have the log tables in front of you for the next series of examples. 1151 They are in Appendix B. 1152 1153 There are four tables. Two are regular log tables. The other two are antilog tables. We’ll start 1154 with the log tables. Looking up logs is relatively easy with these tables. To find the log of 4.25, 1155 we look down the left side of the first table under the column labeled N. We locate the row that 1156 says 4.2, as shown below. 1157


N10log Ten Thousandth Parts N 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

4.1 0.6128 0.6138 0.6149 0.6160 0.6170 0.6180 0.6191 0.6201 0.6212 0.6222 1 2 3 4 5 6 7 8 94.2 0.6232 0.6243 0.6253 0.6263 0.6274 0.6284 0.6294 0.6304 0.6314 0.6325 1 2 3 4 5 6 7 8 94.3 0.6335 0.6345 0.6355 0.6365 0.6375 0.6385 0.6395 0.6405 0.6415 0.6425 1 2 3 4 5 6 7 8 9 1158 Then move along the row until you are in the column that is labeled with the 5. Ignore the “Ten 1159 Thousandth Parts” of the table for now. You should see the number 0.6284. Thus, 25.4log10 = 1160 0.6284. 1161 1162 What does this mean? We go back to the definition of log: 1163 1164

0.628410(0.6284 log 4.25) (10 4.25)= ⇔ = 1165

1166 Hence, the log of 4.25 tells us to what power 10 must be raised to get 4.25. This makes sense 1167 when we remember that 1 < 4.25 < 10. That is: 1168 1169

1 4.25 10< < 1170 So… 0 110 4.25 10< < 1171

1172 The second inequality indicates that the power of 10 we require is somewhere between 0 and 1, 1173 and 0.6284 certainly fits the bill. 1174 1175 Example 17 1176

What is 32.9log10 ? 1177 Solution 1178

Looking at row 9.3 of the table, the number in column 2 is 0.9694. Hence, 1179 100.9694 = 9.32, as desired. This makes sense since 9.32 is close to 10 = 101, so 1180 our log (0.9694) should be close to 1. ♦ 1181

1182 1183 Example 18 1184

What is 4.7log10 ? 1185 Solution 1186

We look up row 7.4 and inside column 0 (since 7.4 = 7.40) we find 0.8692. ♦ 1187 1188 1189 Check Point K 1190

What is 10log 8.55 ? 1191 Solution: 1192

See endnotes for an answer.28 1193 1194 1195


The Ten Thousandth Parts of the tables is there for when the number you are taking the log of 1196 has three decimal places instead of two. To use this part of the table, you take the normal 1197 two−decimal−place log and then add the ten thousandth part to the very end of the result. Here’s 1198 an example: 1199 1200 Example 19 1201

What is log 6.875? 1202 Solution: 1203

For this problem we first compute log 6.87 as before: 1204 N10log Ten Thousandths Parts

N 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 96.7 0.8261 0.8267 0.8274 0.8280 0.8287 0.8293 0.8299 0.8306 0.8312 0.8319 1 1 2 3 3 4 4 5 66.8 0.8325 0.8331 0.8338 0.8344 0.8351 0.8357 0.8363 0.8370 0.8376 0.8382 1 1 2 3 3 4 4 5 66.9 0.8388 0.8395 0.8401 0.8407 0.8414 0.8420 0.8426 0.8432 0.8439 0.8445 1 1 2 2 3 4 4 5 6 1205

log 6.87 = 0.8370. Then, we look under the 5 column of the ten thousandth 1206 parts of the table since the third decimal place of 6.875 is a 5. This gives a 3, 1207 which really represents 0.0003 (That’s why it’s called a proportional part). 1208 Take this and add it to the end of the previous result. So 0.8370+0.0003 = 1209 0.8373. 1210 1211 Hence, log 6.875 = 0.8373. Since our calculators do not exist, we can’t check 1212 this, but rest assured that it’s very close to the actual log of this number. ♦ 1213

1214 Check Point L 1215

Find log 4.638 1216 Solution: 1217

Check the table and make sure you can get 0.6663. 1218 1219

1220 Note that the log requires that numbers strictly between 1 and 10 be used as its inputs. 1221

1222 1223 Larger Numbers and the Log Tables 1224 1225 Okay, what about larger numbers? Suppose we want log 152. The tables don’t have a row with 1226 152 in it, however. They only go up to 9.9. Here we have to rely on some common sense about 1227 powers of 10. We know that 152 > 100, so therefore 152 > 102. Hence, our log must be bigger 1228 than 2 since we need at least 100. We also know that 152 < 103, since 152 < 1000. Thus, 1229 whatever power of 10 that gives us 152 must be between 2 and 3. Hence, log 152 must be 1230 between 2 and 3. 1231 1232 To find how far past 2 we must go, we simply pretend that we are computing log 1.52. When we 1233 look at row 1.5 of the table and move over to column 2, we find 0.1818. That’s how far past two 1234 we must go. Hence, log 152 = 2.1818, since 2.1818 is between 2 and 3. 1235 1236


Here’s another way to think about it: We can rewrite 152 as 100×1.52. Thus, 1237 1238

log152 = 1239 log(100×1.52) = 1240 log100+log1.52. (This last step uses one of the basic log rules.) 1241

1242 We know that log 100 = 2 because 102 = 100. We can get log 1.52 from the table…it’s 0.1818. 1243 So we have log 152=2+0.1818=2.1818 1244 1245 We call 2 (the integer part of the result) the characteristic and we call 0.1818 (the fraction part) 1246 the mantissa. The term characteristic was first used by Briggs. The term mantissa is of Latin 1247 origin, originally meaning an “addition” or “appendix.” It was probably first used by John Wallis 1248 around 1693. 1249 1250 We can streamline some of this by trying to give a series of steps to follow when working with 1251 the log tables. 1252

Step 1 Move the decimal point of the given number so that you have a new number 1253 that is strictly between 1 and 10. 1254

Step 2 Count the number of places that you move the decimal point. This is the 1255 characteristic. 1256

Step 3 Look up the new number (that is between 1 and 10) in the log tables to get the 1257 mantissa. 1258

Step 4 Add the characteristic and the mantissa together to get the final result. 1259 1260 1261 Example 20 1262

Find log 4869. 1263 Solution 1264

If we write this as 4869.0, we note that need to move it 3 places to get 4.869. 1265 Our characteristic, therefore, is 3. (We first note that 1,000<4869<10,000. Put 1266 another way, 103 < 4869 < 104. Hence the power of ten required is more than 1267 3 but less than 4.) To find the mantissa, we look up 4.869 on the table. We 1268 look up row 4.8, column 6, and ten thousandth part 9. We get 0.6866+0.0008 1269 = 0.6874. Combining our characteristic and mantissa gives log 4869 = 3.6874. 1270 Hence, 103.6874 = 4869. ♦ 1271 1272

Check Point M 1273 Find log 35,630. 1274

Solution 1275 Check the endnote for an answer.29 1276

1277


Logs of Very Large Numbers 1278 1279 You have to try to imagine what this means for doing calculations. With these tables, we can 1280 now do sophisticated computations while at the same time getting three or four places of 1281 accuracy without using a calculator (which they did not have at the time anyway). Here’s an 1282 example, however, of why more than three places of accuracy might be needed. 1283 1284 Example 21 1285

Find log 7,834,945. 1286 Solution 1287

We start by finding the characteristic. We need to move the decimal place 6 1288 places to get 7.834945, so our characteristic is 6. To find the mantissa, we find 1289 7.834945 on the log table. Looking up row 7.8, and down column 3 we see 1290 0.8938. But what about the ten thousandth part? Since there are more than 3 1291 decimal places we have to round 7.834945 to 7.835 and look in the 5 column 1292 for the ten thousandth part, which gives us 3, representing 0.0003. Thus, our 1293 mantissa is 0.8938 + 0.0003 = 0.8941. Finally, we get a log of 6+0.8941 = 1294 6.8941. The actual value is something like 6.89403595. Even if we round this 1295 number to the fourth decimal place, we do not get 6.8941. This is due to the 1296 fact that we had to use an estimate to get the ten thousandth part of the 1297 mantissa. ♦ 1298 1299

This may not seem like a big deal to us. After all, three decimal places of accuracy are pretty 1300 good for most applications. But remember that when working with very big or very small 1301 numbers, like astronomers were at that time, losing accuracy in the fourth decimal place could 1302 mean that your calculations end up being significantly off by the time you finish your 1303 calculations. 1304 1305 Eventually, we want to use these tables to do complex calculations but we must first learn to use 1306 the antilog tables. 1307 1308 The Antilog Tables 1309 1310 The analogy of the antilog as the evil twin of the log who undoes what the log does is helpful 1311 here. Whereas the log tables tell us what power to raise 10 to in order to get a result, the antilog 1312 tables tell us the opposite. Hence, the antilog tables are really just power of 10 tables, as you will 1313 see here. 1314 1315 Example 22 1316

If log N = 0.683, what is N? 1317 Solution 1318

Note that this is the opposite question than we had before. Here we know the 1319 log but are looking for the original number N. By the modern definition of 1320 logs, we have: 1321


10

0.683

log 0.683

log 0.683

10

N

N

N

=

=

=

c

c

1322

1323 Hence, we only need to compute 100.683. Without a calculator, we turn to the 1324 antilog tables. 1325 1326 To compute 100.683, we find the 0.68 row and then move over into column 1327 with 3 as its heading. Here is a piece of the antilog table that applies here: 1328 1329

ANTILOG TABLE: 10P = N Thousandth Parts P 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

0.67 4.677 4.688 4.699 4.710 4.721 4.732 4.742 4.753 4.764 4.775 1 2 3 4 5 7 8 9 100.68 4.786 4.797 4.808 4.819 4.831 4.842 4.853 4.864 4.875 4.887 1 2 3 4 6 7 8 9 100.69 4.898 4.909 4.920 4.932 4.943 4.955 4.966 4.977 4.989 5.000 1 2 3 5 6 7 8 9 10

1330 The table indicates that 100.683 = 4.819. This makes some sense when we 1331 remember that since 0<0.683<1, then 100<100.683<101. Hence, we have 1332 1<4.819<10, which is certainly true. ♦ 1333

1334 To use the thousandths parts of the table, we follow the same basic idea as before. 1335 1336 Example 23 1337

If log N = 0.8235, find N. 1338 Solution 1339

By the definition of log, we have 100.8235=N. We first find row 0.82 and then 1340 move over to column 3. This takes us to 6.653. 1341 1342


0.81 6.457 6.471 6.486 6.501 6.516 6.531 6.546 6.561 6.577 6.592 2 3 5 6 8 9 11 12 140.82 6.607 6.622 6.637 6.653 6.668 6.683 6.699 6.714 6.730 6.745 2 3 5 6 8 9 11 12 140.83 6.761 6.776 6.792 6.808 6.823 6.839 6.855 6.871 6.887 6.902 2 3 5 6 8 10 11 13 14

1343 Then, we look up the 5 in the thousandth parts of the table and see and 8 there, 1344 which now represents 0.008. We add these, 6.653+0.008 = 6.661. Therefore, 1345 we have 100.8235 = 6.661. ♦ 1346 1347

1348 Note that the antilog requires numbers between 0 and 1 as inputs. 1349

1350 1351


Check Point N 1352 If logN = 0.1834, find N. What does N represent as a power of 10? 1353

Solution 1354 See the endnote for an answer.30 1355

1356 1357 So then what do we do about larger numbers? 1358 1359 1360 Example 24 1361

If logN = 3.874, what is N? 1362 Solution 1363

We rewrite this as 103.874 = N. 1364 First we notice that 103.874 = 103 × 100.874. (Rules of exponents… )nmnm xxx =+ 1365 Thus, 103.874 = 1000 × 100.874. 1366 1367 The power of 10 on the end can be obtained from the table. 100.874 = 7.482. 1368 So we have 1000×7.482 = 7482. 1369 1370 The actual value is 7481.69500511, so we are reasonably close (four places of 1371 accuracy when rounded.). ♦ 1372

1373 Example 25 1374

If logN = 7.2159, find N. 1375 Solution 1376

First we write 107.2159 = N. 1377 Next, we write N = 107 × 100.2159 = 10,000,000 × 100.2159 1378 From the table we have: N = 10,000,000 × 1.644 = 16,440,000. ♦ 1379

1380 There is an alternative way to approach antilog problems. Notice that in the previous few 1381 examples, the whole number part of the number we are given does not play a role when we use 1382 the tables. Only the fractional part has an impact. So, as a shortcut, we can look up the fractional 1383 part in the antilog tables and then move the decimal point in the result. The number of places that 1384 we move corresponds to the whole number part of the given number. In the previous example, 1385 note that you had to move the decimal point 7 places which corresponds to the 7 in 7.2159. 1386 1387 1388 Check Point O 1389

If logN = 5.623, find N. 1390 Solution 1391

See the endnotes to check your answer.31 1392 1393

1394


The Evil Twins Meet Each Other…Finally 1395 1396 We have been saying that antilogs and logs undo each other’s work. Let’s see a pair of examples 1397 that demonstrate that. 1398 1399 Example 26 1400

What is the value of x = log 76,540? 1401 Solution 1402

We note that 104 < 76,540 < 105. This means our characteristic is 4 and we 1403 need to find the mantissa from the table. Since the table requires an input 1404 between 1 and 10, we use 7.654. From the log table, we get 1405 0.8837+0.0002=0.8839. Hence, log 76,540 = 4.8839. This means that 104.8839 1406 = 76540. ♦ 1407

1408 Now we’ll do the opposite of this problem…it’s twin. 1409 1410 Example 27 1411

If logN = 4.8839, what is N? 1412 Solution 1413

This is linked to the previous example…we should expect to get back 76,540. 1414 (Why?) Because logN = 4.8838 means that 104.8838 = N, we can use the 1415 antilog table to find N. 1416 1417 104.8838 = 104×100.8838 = 10,000×100.8838 = 10,000×(7.638+.014) = 1418 10,000×7.652 = 76,520. 1419 1420 Okay….we’re off a little bit, but if we had more decimal places of accuracy in 1421 our table, we would get an even better result. ♦ 1422 1423

What is important to observe from the last two examples is that applying log and antilog tables 1424 basically take you back to the original starting point. For this reason, we say that the log and 1425 antilog tables are inverses of each other in the same way that square roots and powers of two are 1426 inverses of each other. 1427

1428 1429

Complex Calculations 1430 1431 Now imagine for a moment that you are living in the 1600’s and your astronomical calculations 1432 require that you multiply 874,200,000×7,853,00. You have no calculator. You don’t have any 1433 easy method of multiplying these two numbers…until Napier knocks on your door and 1434 introduces you to his logs. With our three basic log rules, and our four handy log/antilog tables, 1435 we can do this and many other nasty computations. 1436


Recall: 1437 Definition of log 1438

nbpn pb =⇔=log 1439

1440 Basic log Rules 1441

ncn

BABA

BAAB

c loglog.)3

logloglog.)2

logloglog.)1

=

−=

+=

1442

1443 1444 Example 28 1445

Multiply 874,200,000 × 7,853,000 1446 Solution 1447

We’ll start all such problems by letting x equal the quantity we want to 1448 compute. 1449 1450

Operation Comments x = 874,200,000×7,853,000 This is our computation

7,853,000)0874,200,00log(log ×=x Take the log of both sides. This is going to transform our statement into one with logs, taking us into “logland.” It’s a key step!

000,853,7log000,200,874loglog +=x Apply log rule #1 (see above) 108 < 874,200,000 < 109, so

9416.8742.8log8000,200,874log =+= Compute log 874,200,000 with the table and save it for later. You can also think of 874,200,000 as 8.742×108 to help find the characteristic.

106 < 7,853,000 < 107, so 8951.6853.7log6000,853,7log =+=

Compute log 7,853,000 with the table. You can also think of 7,853,000 as 7.853×106 to help find the characteristic.

log x = 8.9416 + 6.8951 = 15.8367 Add these two logs together by hand. Addition is easy! This is not our answer, of course. We now need to call in the evil twin to take us out of logland.

xx =⇔= 8367.15108367.15log Rewrite using the definition of logs

866.610101010 158367.158367.15 ×=×==x Use the antilog table to compute what x is. Remember, x is the desired quantity.

x = 6.866×1015 We’ll leave this in scientific notation. 1451


If we cheat and take out our calculator, we’ll find that 874,200,000×7,853,00 1452 = 6.8650926×1015. This means that our log table was good to three decimal 1453 places, but missed the fourth. With tables that had more accuracy, we could 1454 get a better answer. ♦ 1455

1456 1457 Check Point P 1458

Multiply 357,600,000×4,591,000,000,000 using the log and antilog tables. 1459 Solution 1460

See the endnotes to check your answer.32 1461 1462 1463 Example 29 1464

Compute 028.2

5.893=x with the log tables. 1465

Solution 1466 Although these numbers are not very ugly, they will demonstrate the process 1467 just fine. 1468 1469

Operation Comments

028.25.893

=x

=

028.25.893loglog x

Take the log of both sides

028.2log5.893loglog −=x Apply log rule #2 644.23071.09511.2log =−=x Use the log tables to carefully find the two logs on the

right. You should verify these before you proceed. Note that the characteristic of log893.5 is 2 since 102<893.5<103. The characteristic of 2.028 is 0. (Why?)

xx =⇔= 644.102644.2log Definition of log

406.41010100101010 644.0644.02644.2

×=×=×==x

Rewrite and use the antilog table to compute the value of x.

x = 440.6 This is our final answer. A calculator gives 440.555 as the final answer. Not too bad (but not that great, either).♦

1470 Check Point Q 1471

Compute 235.6017.22

x = with the log tables. 1472

Solution 1473 See endnotes for a solution.33 1474

1475


Example 30 1476 Find the value of y = 84.658 1477

Solution 1478 Many complex important calculations require that numbers be raised to 1479 powers like this. The tables can help us here as well. 1480 1481

Operation Comments y = 84.658

865.84loglog =y Take the log of both sides. 65.84log8log =y Use log rule #3 to move the power in front of the

log. 9277.18)9277.01(8log ×=+×=y Take the log of 84.65. Note that the characteristic is

1 since 101<84.65<102. At this point we have to do 8×1.9277. While we could use a log table to do this (nested logs, yuck), we observe that a simple multiplication like this is probably something Napier’s contemporaries could have done…they were concerned about the nasty ones.

4216.15log =y Multiply by hand. 8×1.9277. Now we know what log y is. We can undo the log by using the antilog table.

4216.0154216.15 1010104216.15log ×=⇔=y Apply the log definition and simplify so that we can use the antilog table.

1515 1064.2640.210log ×=×=y Use the antilog table to do 100.4216 and we have a final answer. A calculator gives the result to be 2.63643×1015.

1482 This last example truly shows the power of logarithms. In the 1600’s, it was 1483 not easy at all to do this last kind of computation. By converting to logs, 1484 working with addition and subtraction (and perhaps some easy multiplication), 1485 and then converting back from logs to regular numbers, they could raise a 1486 decimal fraction to a power with relatively little work. ♦ 1487

1488 Check Point R 1489

Compute 278.410 with the log tables. 1490 Solution 1491

See endnotes for a solution.34 1492 1493 We can even take roots of numbers with logs. We simply need to remember a basic rule of 1494 exponents: 1495 1496

n mnm xx =/ 1497 1498 1499


This tells us how to express roots with fractional powers. Here are two common identities: 1500 1501

2/1xx = 1502 3/13 xx = 1503

1504 1505 Example 31 1506

Find the value of 2.789=x 1507 Solution 1508 1509 Operation Comments

( ) 2/12.7892.789 ==x Use the exponent rule to convert.

( ) 2/12.789loglog =x Take the log of both sides.

2.789log21log =x Use log rule #3 to move the power in front.

( )8972.221log =x Use the log table to compute log 789.2

4486.1log =x Take half of the result by hand…this is easy enough to do without invoking another layer of logs.

09.28809.210101010 4486.01446.1 =×=×==x Rewrite x and then use the antilog table to compute 100.4486. The final answer of 28.09 compares with a calculator result of 28.0931. ♦

1510 1511 Check Point S 1512

Find the value of 3 1,342,000 using the log tables 1513 Solution: 1514

See endnotes for a solution.35 1515 1516 With this capability in hand, astronomers, mathematicians, and other scientists could speed up 1517 their calculations and accelerate their discoveries. Of course, they still had to learn the basic 1518 skills of arithmetic, but that wasn’t a major problem. It is analogous to when the handheld 1519 calculator infiltrated the classroom. They made calculations much faster and easier, but they are 1520 not the answer to learning math. Students of math still need to learn the basics so that they can 1521 check the reasonableness of answers and so that when confronted with a relatively simple 1522 problem, they don’t have to go searching for a calculator to pound on. Unfortunately, it seems as 1523 though we are overly dependent on calculators. For some, it slows them down since every 1524 computation must be done with that wretched piece of plastic. (I’ve even seen students reach for 1525 their calculator when asked what 2×3 is!). For most, it leads to a false sense of security about 1526 “answers” because their addiction to the calculating machine leaves them powerless to question 1527 whether or not their answers even make sense. 1528


PART 5: New Calculating Devices 1529 1530 The Slide Rule 1531 1532 As word of the logarithm spread, mathematicians took the idea and not only used it to help them 1533 with their work, but also extended it. Edmund Gunter, a faculty member at Gresham College 1534 where Briggs was a professor of geometry, invented a tool with log, tangent, and sine scales 1535 which was used by navigators. In 1621, William Oughtred invented a slide rule that helped with 1536 computations. The slide rule slowly evolved and became an important computing device until the 1537 handheld calculator and desktop computer emerged as the computing workhorses of the modern 1538 era. 1539 1540 Napier and Other Calculating Devices 1541 1542 Napier is also known for creating another, less−powerful computing device called “Napier’s 1543 Bones.” They are multiplication devices based on the gelosia method of multiplication that we 1544 discussed earlier. Napier noted that the columns on the gelosia grid were merely multiples of 1545 numbers at the top of the columns. He took the columns and placed them on vertical “rods” 1546 which could then be moved around easily and used as a mobile computing device. He called this 1547 invention “Rabdologia” which means “a collection of rods” in Greek. They eventually picked up 1548 the nickname “Napier’s Rods,” or “Napier’s Bones.” There is a full set of the rods in the 1549 appendix that you can cut out if you would like. We can see how these rods 1550 work with an example. Let’s compute 486×7. To do this: 1551 1552 1. We take the 4, 8, and 6 rods and place them in order next to each other, 1553

and we place the “index rod” on the right of these. 1554 2. To find the product 486×7, read along the row marked with a 7 on the 1555

index rod. 1556 3. Add the numbers along the diagonals, just as before, to get 3402. 1557

Ignore all other rows, as they are not needed (since we are multiplying 1558 by 7). 1559

1560 1561 1562 4 1563 1564 As you can imagine, this eliminates the need to write the grid down because 1565 you can rearrange rods and read off results rather quickly. 1566 1567 Example 32 1568

Use the rods to find the product 569×27 1569 Solution 1570

Here we have two steps to take. First multiply 569 by 7. 1571 Next, multiply 569 by 2 tens. Essentially, we are thinking 1572 of this problem as (569)×(7×1)+569×(2×10) 1573

0 0 04 8 6

0 1 18 6 2

1 2 12 4 8

1 3 26 2 4

2 4 30 0 0

2 4 34 8 6

2 5 48 6 2

3 6 42 4 8

3 7 56 2 4

4 8 6

1

2

3

4

9

5

6

7

8

0 0 05 6 9

1 1 10 2 8

1 1 25 8 7

2 2 30 4 6

2 3 45 0 5

3 3 60 6 4

3 4 65 2 3

4 4 70 8 2

4 5 85 4 1

1

5 6 9

2

3

4

5

6

7

8

9

2 5 48 6 23

04 2


569×(7×ones) = (569×7)×1 = 3983×1 = 03983 1574 569×(2×tens) = (569×2)×10 = 1138×10 = 11380 1575 Result = 15363 1576 1577 1578 We are once again thinking of multiplication in terms of its relationship to 1579 addition. By adding together all the appropriate places, we get to a result.♦ 1580

1581 Example 33 1582

Use the rods to compute 631×854. 1583 Solution 1584

We think of this as 631×(4×1) + 631×(5×10) + 1585 631×(8×100) 1586

631×(4×1) = (631×4)×1 = 2524 1587 631×(5×10) = (631×5)×10 = 31550 1588 631×(8×100) = (631×8)×100 = 504800 1589 Result = 538,874 1590 1591

A calculator check will show this to be true.♦ 1592 1593

Since addition was (and is) much easier than multiplication, this method 1594 provided a nice way to be able to compute “on the fly.” In a time when 1595 merchants and other professionals in fields such as commerce were required 1596 to do increasingly complication computations, time−saving devices like 1597 these were often cherished. It is reported that some people even “jealousy” guarded their secrets 1598 from others so that they could have a computational advantage. 1599 1600 Check Point T 1601

Use your own set of the rods to find 542×8, and 986×36 1602 Solution 1603

You can use a calculator to check your answers. 1604

0 0 06 3 1

1 0 02 6 2

1 0 08 9 3

2 1 04 2 4

3 1 00 5 5

3 1 06 8 6

4 2 02 1 7

4 2 08 4 8

5 2 04 7 9

6

7

8

9

2

3

4

5

6 3 1

1

Think About It

Could you adapt the rods so that

they would compute

785×65.3? How?


The rods were eventually manufactured and carried around so that they could be used to 1605 “mobile” calculating. Here are some sets of Napier rods that show how they looked.36 1606 1607

1608 1609


1610 PART 6: Appendix A − Solution of the Quartic37 1611 A quartic equation of the form: 1612 1613

0234 =++++ dcxbxaxx 1614 1615

has a related equation (its resolvent cubic equation) of the form: 1616 1617

04)4( 2223 =−+−−+− cbddaydacbyy 1618 1619

Let y be any solution of this equation and let 1620 1621

ybaR +−=4

2

1622

1623 If R is not 0, then let 1624 1625 1626

1627 and 1628

RacabbRaE

4842

43 3

22 −−

−−−= 1629

1630 If R = 0 then let 1631

dybaD 4224

3 22

−+−= 1632

and 1633

dybaE 4224

3 22

−−−= 1634

1635 Then the four solutions of the original equation are given by: 1636 1637

224DRax ±+−= 1638

224ERax ±−−= 1639

1640 Holy Cow!!! 1641

2 323 4 82

4 4a ab c aD R b

R− −

= − − +


Blank page1642


PART 7: Appendix B – Log and Antilog Tables 1643 N10log Ten Thousandth Parts

N 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 91.0 0.0000 0.0043 0.0086 0.0128 0.0170 0.0212 0.0253 0.0294 0.0334 0.0374 4 8 12 16 20 24 28 32 361.1 0.0414 0.0453 0.0492 0.0531 0.0569 0.0607 0.0645 0.0682 0.0719 0.0755 4 7 11 15 18 22 26 29 331.2 0.0792 0.0828 0.0864 0.0899 0.0934 0.0969 0.1004 0.1038 0.1072 0.1106 3 7 10 14 17 20 24 27 301.3 0.1139 0.1173 0.1206 0.1239 0.1271 0.1303 0.1335 0.1367 0.1399 0.1430 3 6 9 13 16 19 22 25 281.4 0.1461 0.1492 0.1523 0.1553 0.1584 0.1614 0.1644 0.1673 0.1703 0.1732 3 6 9 12 15 18 20 23 261.5 0.1761 0.1790 0.1818 0.1847 0.1875 0.1903 0.1931 0.1959 0.1987 0.2014 3 5 8 11 14 16 19 22 251.6 0.2041 0.2068 0.2095 0.2122 0.2148 0.2175 0.2201 0.2227 0.2253 0.2279 3 5 8 10 13 15 18 21 231.7 0.2304 0.2330 0.2355 0.2380 0.2405 0.2430 0.2455 0.2480 0.2504 0.2529 2 5 7 10 12 15 17 19 221.8 0.2553 0.2577 0.2601 0.2625 0.2648 0.2672 0.2695 0.2718 0.2742 0.2765 2 5 7 9 12 14 16 18 211.9 0.2788 0.2810 0.2833 0.2856 0.2878 0.2900 0.2923 0.2945 0.2967 0.2989 2 4 7 9 11 13 15 18 202.0 0.3010 0.3032 0.3054 0.3075 0.3096 0.3118 0.3139 0.3160 0.3181 0.3201 2 4 6 8 10 12 15 17 192.1 0.3222 0.3243 0.3263 0.3284 0.3304 0.3324 0.3345 0.3365 0.3385 0.3404 2 4 6 8 10 12 14 16 182.2 0.3424 0.3444 0.3464 0.3483 0.3502 0.3522 0.3541 0.3560 0.3579 0.3598 2 4 6 8 10 11 13 15 172.3 0.3617 0.3636 0.3655 0.3674 0.3692 0.3711 0.3729 0.3747 0.3766 0.3784 2 4 5 7 9 11 13 15 162.4 0.3802 0.3820 0.3838 0.3856 0.3874 0.3892 0.3909 0.3927 0.3945 0.3962 2 3 5 7 9 10 12 14 162.5 0.3979 0.3997 0.4014 0.4031 0.4048 0.4065 0.4082 0.4099 0.4116 0.4133 2 3 5 7 8 10 12 13 152.6 0.4150 0.4166 0.4183 0.4200 0.4216 0.4232 0.4249 0.4265 0.4281 0.4298 2 3 5 6 8 10 11 13 152.7 0.4314 0.4330 0.4346 0.4362 0.4378 0.4393 0.4409 0.4425 0.4440 0.4456 2 3 5 6 8 9 11 12 142.8 0.4472 0.4487 0.4502 0.4518 0.4533 0.4548 0.4564 0.4579 0.4594 0.4609 2 3 5 6 8 9 11 12 142.9 0.4624 0.4639 0.4654 0.4669 0.4683 0.4698 0.4713 0.4728 0.4742 0.4757 1 3 4 6 7 9 10 12 133.0 0.4771 0.4786 0.4800 0.4814 0.4829 0.4843 0.4857 0.4871 0.4886 0.4900 1 3 4 6 7 8 10 11 133.1 0.4914 0.4928 0.4942 0.4955 0.4969 0.4983 0.4997 0.5011 0.5024 0.5038 1 3 4 5 7 8 10 11 123.2 0.5051 0.5065 0.5079 0.5092 0.5105 0.5119 0.5132 0.5145 0.5159 0.5172 1 3 4 5 7 8 9 11 123.3 0.5185 0.5198 0.5211 0.5224 0.5237 0.5250 0.5263 0.5276 0.5289 0.5302 1 3 4 5 6 8 9 10 123.4 0.5315 0.5328 0.5340 0.5353 0.5366 0.5378 0.5391 0.5403 0.5416 0.5428 1 2 4 5 6 7 9 10 113.5 0.5441 0.5453 0.5465 0.5478 0.5490 0.5502 0.5514 0.5527 0.5539 0.5551 1 2 4 5 6 7 8 10 113.6 0.5563 0.5575 0.5587 0.5599 0.5611 0.5623 0.5635 0.5647 0.5658 0.5670 1 2 4 5 6 7 8 9 113.7 0.5682 0.5694 0.5705 0.5717 0.5729 0.5740 0.5752 0.5763 0.5775 0.5786 1 2 3 5 6 7 8 9 103.8 0.5798 0.5809 0.5821 0.5832 0.5843 0.5855 0.5866 0.5877 0.5888 0.5899 1 2 3 4 6 7 8 9 103.9 0.5911 0.5922 0.5933 0.5944 0.5955 0.5966 0.5977 0.5988 0.5999 0.6010 1 2 3 4 5 7 8 9 104.0 0.6021 0.6031 0.6042 0.6053 0.6064 0.6075 0.6085 0.6096 0.6107 0.6117 1 2 3 4 5 6 7 9 104.1 0.6128 0.6138 0.6149 0.6160 0.6170 0.6180 0.6191 0.6201 0.6212 0.6222 1 2 3 4 5 6 7 8 94.2 0.6232 0.6243 0.6253 0.6263 0.6274 0.6284 0.6294 0.6304 0.6314 0.6325 1 2 3 4 5 6 7 8 94.3 0.6335 0.6345 0.6355 0.6365 0.6375 0.6385 0.6395 0.6405 0.6415 0.6425 1 2 3 4 5 6 7 8 94.4 0.6435 0.6444 0.6454 0.6464 0.6474 0.6484 0.6493 0.6503 0.6513 0.6522 1 2 3 4 5 6 7 8 94.5 0.6532 0.6542 0.6551 0.6561 0.6571 0.6580 0.6590 0.6599 0.6609 0.6618 1 2 3 4 5 6 7 8 94.6 0.6628 0.6637 0.6646 0.6656 0.6665 0.6675 0.6684 0.6693 0.6702 0.6712 1 2 3 4 5 6 6 7 84.7 0.6721 0.6730 0.6739 0.6749 0.6758 0.6767 0.6776 0.6785 0.6794 0.6803 1 2 3 4 5 5 6 7 84.8 0.6812 0.6821 0.6830 0.6839 0.6848 0.6857 0.6866 0.6875 0.6884 0.6893 1 2 3 4 4 5 6 7 84.9 0.6902 0.6911 0.6920 0.6928 0.6937 0.6946 0.6955 0.6964 0.6972 0.6981 1 2 3 3 4 5 6 7 85.0 0.6990 0.6998 0.7007 0.7016 0.7024 0.7033 0.7042 0.7050 0.7059 0.7067 1 2 3 3 4 5 6 7 8

1644


1645 N10log Ten Thousandth Parts

N 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 5.1 0.7076 0.7084 0.7093 0.7101 0.7110 0.7118 0.7126 0.7135 0.7143 0.7152 1 2 3 3 4 5 6 7 8 5.2 0.7160 0.7168 0.7177 0.7185 0.7193 0.7202 0.7210 0.7218 0.7226 0.7235 1 2 2 3 4 5 6 7 7 5.3 0.7243 0.7251 0.7259 0.7267 0.7275 0.7284 0.7292 0.7300 0.7308 0.7316 1 2 2 3 4 5 6 6 7 5.4 0.7324 0.7332 0.7340 0.7348 0.7356 0.7364 0.7372 0.7380 0.7388 0.7396 1 2 2 3 4 5 6 6 7 5.5 0.7404 0.7412 0.7419 0.7427 0.7435 0.7443 0.7451 0.7459 0.7466 0.7474 1 2 2 3 4 5 5 6 7 5.6 0.7482 0.7490 0.7497 0.7505 0.7513 0.7520 0.7528 0.7536 0.7543 0.7551 1 2 2 3 4 5 5 6 7 5.7 0.7559 0.7566 0.7574 0.7582 0.7589 0.7597 0.7604 0.7612 0.7619 0.7627 1 2 2 3 4 5 5 6 7 5.8 0.7634 0.7642 0.7649 0.7657 0.7664 0.7672 0.7679 0.7686 0.7694 0.7701 1 1 2 3 4 4 5 6 7 5.9 0.7709 0.7716 0.7723 0.7731 0.7738 0.7745 0.7752 0.7760 0.7767 0.7774 1 1 2 3 4 4 5 6 7 6.0 0.7782 0.7789 0.7796 0.7803 0.7810 0.7818 0.7825 0.7832 0.7839 0.7846 1 1 2 3 4 4 5 6 6 6.1 0.7853 0.7860 0.7868 0.7875 0.7882 0.7889 0.7896 0.7903 0.7910 0.7917 1 1 2 3 4 4 5 6 6 6.2 0.7924 0.7931 0.7938 0.7945 0.7952 0.7959 0.7966 0.7973 0.7980 0.7987 1 1 2 3 3 4 5 6 6 6.3 0.7993 0.8000 0.8007 0.8014 0.8021 0.8028 0.8035 0.8041 0.8048 0.8055 1 1 2 3 3 4 5 5 6 6.4 0.8062 0.8069 0.8075 0.8082 0.8089 0.8096 0.8102 0.8109 0.8116 0.8122 1 1 2 3 3 4 5 5 6 6.5 0.8129 0.8136 0.8142 0.8149 0.8156 0.8162 0.8169 0.8176 0.8182 0.8189 1 1 2 3 3 4 5 5 6 6.6 0.8195 0.8202 0.8209 0.8215 0.8222 0.8228 0.8235 0.8241 0.8248 0.8254 1 1 2 3 3 4 5 5 6 6.7 0.8261 0.8267 0.8274 0.8280 0.8287 0.8293 0.8299 0.8306 0.8312 0.8319 1 1 2 3 3 4 4 5 6 6.8 0.8325 0.8331 0.8338 0.8344 0.8351 0.8357 0.8363 0.8370 0.8376 0.8382 1 1 2 3 3 4 4 5 6 6.9 0.8388 0.8395 0.8401 0.8407 0.8414 0.8420 0.8426 0.8432 0.8439 0.8445 1 1 2 2 3 4 4 5 6 7.0 0.8451 0.8457 0.8463 0.8470 0.8476 0.8482 0.8488 0.8494 0.8500 0.8506 1 1 2 2 3 4 4 5 6 7.1 0.8513 0.8519 0.8525 0.8531 0.8537 0.8543 0.8549 0.8555 0.8561 0.8567 1 1 2 2 3 4 4 5 5 7.2 0.8573 0.8579 0.8585 0.8591 0.8597 0.8603 0.8609 0.8615 0.8621 0.8627 1 1 2 2 3 4 4 5 5 7.3 0.8633 0.8639 0.8645 0.8651 0.8657 0.8663 0.8669 0.8675 0.8681 0.8686 1 1 2 2 3 4 4 5 5 7.4 0.8692 0.8698 0.8704 0.8710 0.8716 0.8722 0.8727 0.8733 0.8739 0.8745 1 1 2 2 3 3 4 5 5 7.5 0.8751 0.8756 0.8762 0.8768 0.8774 0.8779 0.8785 0.8791 0.8797 0.8802 1 1 2 2 3 3 4 5 5 7.6 0.8808 0.8814 0.8820 0.8825 0.8831 0.8837 0.8842 0.8848 0.8854 0.8859 1 1 2 2 3 3 4 5 5 7.7 0.8865 0.8871 0.8876 0.8882 0.8887 0.8893 0.8899 0.8904 0.8910 0.8915 1 1 2 2 3 3 4 4 5 7.8 0.8921 0.8927 0.8932 0.8938 0.8943 0.8949 0.8954 0.8960 0.8965 0.8971 1 1 2 2 3 3 4 4 5 7.9 0.8976 0.8982 0.8987 0.8993 0.8998 0.9004 0.9009 0.9015 0.9020 0.9025 1 1 2 2 3 3 4 4 5 8.0 0.9031 0.9036 0.9042 0.9047 0.9053 0.9058 0.9063 0.9069 0.9074 0.9079 1 1 2 2 3 3 4 4 5 8.1 0.9085 0.9090 0.9096 0.9101 0.9106 0.9112 0.9117 0.9122 0.9128 0.9133 1 1 2 2 3 3 4 4 5 8.2 0.9138 0.9143 0.9149 0.9154 0.9159 0.9165 0.9170 0.9175 0.9180 0.9186 1 1 2 2 3 3 4 4 5 8.3 0.9191 0.9196 0.9201 0.9206 0.9212 0.9217 0.9222 0.9227 0.9232 0.9238 1 1 2 2 3 3 4 4 5 8.4 0.9243 0.9248 0.9253 0.9258 0.9263 0.9269 0.9274 0.9279 0.9284 0.9289 1 1 2 2 3 3 4 4 5 8.5 0.9294 0.9299 0.9304 0.9309 0.9315 0.9320 0.9325 0.9330 0.9335 0.9340 1 1 2 2 3 3 4 4 5 8.6 0.9345 0.9350 0.9355 0.9360 0.9365 0.9370 0.9375 0.9380 0.9385 0.9390 1 1 2 2 3 3 4 4 5 8.7 0.9395 0.9400 0.9405 0.9410 0.9415 0.9420 0.9425 0.9430 0.9435 0.9440 0 1 1 2 2 3 3 4 4 8.8 0.9445 0.9450 0.9455 0.9460 0.9465 0.9469 0.9474 0.9479 0.9484 0.9489 0 1 1 2 2 3 3 4 4 8.9 0.9494 0.9499 0.9504 0.9509 0.9513 0.9518 0.9523 0.9528 0.9533 0.9538 0 1 1 2 2 3 3 4 4 9.0 0.9542 0.9547 0.9552 0.9557 0.9562 0.9566 0.9571 0.9576 0.9581 0.9586 0 1 1 2 2 3 3 4 4 9.1 0.9590 0.9595 0.9600 0.9605 0.9609 0.9614 0.9619 0.9624 0.9628 0.9633 0 1 1 2 2 3 3 4 4 9.2 0.9638 0.9643 0.9647 0.9652 0.9657 0.9661 0.9666 0.9671 0.9675 0.9680 0 1 1 2 2 3 3 4 4 9.3 0.9685 0.9689 0.9694 0.9699 0.9703 0.9708 0.9713 0.9717 0.9722 0.9727 0 1 1 2 2 3 3 4 4 9.4 0.9731 0.9736 0.9741 0.9745 0.9750 0.9754 0.9759 0.9763 0.9768 0.9773 0 1 1 2 2 3 3 4 4 9.5 0.9777 0.9782 0.9786 0.9791 0.9795 0.9800 0.9805 0.9809 0.9814 0.9818 0 1 1 2 2 3 3 4 4 9.6 0.9823 0.9827 0.9832 0.9836 0.9841 0.9845 0.9850 0.9854 0.9859 0.9863 0 1 1 2 2 3 3 4 4 9.7 0.9868 0.9872 0.9877 0.9881 0.9886 0.9890 0.9894 0.9899 0.9903 0.9908 0 1 1 2 2 3 3 4 4 9.8 0.9912 0.9917 0.9921 0.9926 0.9930 0.9934 0.9939 0.9943 0.9948 0.9952 0 1 1 2 2 3 3 4 4 9.9 0.9956 0.9961 0.9965 0.9969 0.9974 0.9978 0.9983 0.9987 0.9991 0.9996 0 1 1 2 2 3 3 3 4

1646



0.00 1.000 1.002 1.005 1.007 1.009 1.012 1.014 1.016 1.019 1.021 0 0 1 1 1 1 2 2 2 0.01 1.023 1.026 1.028 1.030 1.033 1.035 1.038 1.040 1.042 1.045 0 0 1 1 1 1 2 2 2 0.02 1.047 1.050 1.052 1.054 1.057 1.059 1.062 1.064 1.067 1.069 0 0 1 1 1 1 2 2 2 0.03 1.072 1.074 1.076 1.079 1.081 1.084 1.086 1.089 1.091 1.094 0 1 1 1 1 2 2 2 2 0.04 1.096 1.099 1.102 1.104 1.107 1.109 1.112 1.114 1.117 1.119 0 1 1 1 1 2 2 2 2 0.05 1.122 1.125 1.127 1.130 1.132 1.135 1.138 1.140 1.143 1.146 0 1 1 1 1 2 2 2 2 0.06 1.148 1.151 1.153 1.156 1.159 1.161 1.164 1.167 1.169 1.172 0 1 1 1 1 2 2 2 2 0.07 1.175 1.178 1.180 1.183 1.186 1.189 1.191 1.194 1.197 1.199 0 1 1 1 1 2 2 2 2 0.08 1.202 1.205 1.208 1.211 1.213 1.216 1.219 1.222 1.225 1.227 0 1 1 1 1 2 2 2 3 0.09 1.230 1.233 1.236 1.239 1.242 1.245 1.247 1.250 1.253 1.256 0 1 1 1 1 2 2 2 3 0.10 1.259 1.262 1.265 1.268 1.271 1.274 1.276 1.279 1.282 1.285 0 1 1 1 1 2 2 2 3 0.11 1.288 1.291 1.294 1.297 1.300 1.303 1.306 1.309 1.312 1.315 0 1 1 1 2 2 2 2 3 0.12 1.318 1.321 1.324 1.327 1.330 1.334 1.337 1.340 1.343 1.346 0 1 1 1 2 2 2 2 3 0.13 1.349 1.352 1.355 1.358 1.361 1.365 1.368 1.371 1.374 1.377 0 1 1 1 2 2 2 3 3 0.14 1.380 1.384 1.387 1.390 1.393 1.396 1.400 1.403 1.406 1.409 0 1 1 1 2 2 2 3 3 0.15 1.413 1.416 1.419 1.422 1.426 1.429 1.432 1.435 1.439 1.442 0 1 1 1 2 2 2 3 3 0.16 1.445 1.449 1.452 1.455 1.459 1.462 1.466 1.469 1.472 1.476 0 1 1 1 2 2 2 3 3 0.17 1.479 1.483 1.486 1.489 1.493 1.496 1.500 1.503 1.507 1.510 0 1 1 1 2 2 2 3 3 0.18 1.514 1.517 1.521 1.524 1.528 1.531 1.535 1.538 1.542 1.545 0 1 1 1 2 2 2 3 3 0.19 1.549 1.552 1.556 1.560 1.563 1.567 1.570 1.574 1.578 1.581 0 1 1 1 2 2 3 3 3 0.20 1.585 1.589 1.592 1.596 1.600 1.603 1.607 1.611 1.614 1.618 0 1 1 1 2 2 3 3 3 0.21 1.622 1.626 1.629 1.633 1.637 1.641 1.644 1.648 1.652 1.656 0 1 1 2 2 2 3 3 3 0.22 1.660 1.663 1.667 1.671 1.675 1.679 1.683 1.687 1.690 1.694 0 1 1 2 2 2 3 3 4 0.23 1.698 1.702 1.706 1.710 1.714 1.718 1.722 1.726 1.730 1.734 0 1 1 2 2 2 3 3 4 0.24 1.738 1.742 1.746 1.750 1.754 1.758 1.762 1.766 1.770 1.774 0 1 1 2 2 2 3 3 4 0.25 1.778 1.782 1.786 1.791 1.795 1.799 1.803 1.807 1.811 1.816 0 1 1 2 2 3 3 3 4 0.26 1.820 1.824 1.828 1.832 1.837 1.841 1.845 1.849 1.854 1.858 0 1 1 2 2 3 3 3 4 0.27 1.862 1.866 1.871 1.875 1.879 1.884 1.888 1.892 1.897 1.901 0 1 1 2 2 3 3 3 4 0.28 1.905 1.910 1.914 1.919 1.923 1.928 1.932 1.936 1.941 1.945 0 1 1 2 2 3 3 4 4 0.29 1.950 1.954 1.959 1.963 1.968 1.972 1.977 1.982 1.986 1.991 0 1 1 2 2 3 3 4 4 0.30 1.995 2.000 2.004 2.009 2.014 2.018 2.023 2.028 2.032 2.037 0 1 1 2 2 3 3 4 4 0.31 2.042 2.046 2.051 2.056 2.061 2.065 2.070 2.075 2.080 2.084 0 1 1 2 2 3 3 4 4 0.32 2.089 2.094 2.099 2.104 2.109 2.113 2.118 2.123 2.128 2.133 0 1 1 2 2 3 3 4 4 0.33 2.138 2.143 2.148 2.153 2.158 2.163 2.168 2.173 2.178 2.183 1 1 2 2 3 3 4 4 5 0.34 2.188 2.193 2.198 2.203 2.208 2.213 2.218 2.223 2.228 2.234 1 1 2 2 3 3 4 4 5 0.35 2.239 2.244 2.249 2.254 2.259 2.265 2.270 2.275 2.280 2.286 1 1 2 2 3 3 4 4 5 0.36 2.291 2.296 2.301 2.307 2.312 2.317 2.323 2.328 2.333 2.339 1 1 2 2 3 3 4 4 5 0.37 2.344 2.350 2.355 2.360 2.366 2.371 2.377 2.382 2.388 2.393 1 1 2 2 3 3 4 4 5 0.38 2.399 2.404 2.410 2.415 2.421 2.427 2.432 2.438 2.443 2.449 1 1 2 2 3 3 4 5 5 0.39 2.455 2.460 2.466 2.472 2.477 2.483 2.489 2.495 2.500 2.506 1 1 2 2 3 3 4 5 5 0.40 2.512 2.518 2.523 2.529 2.535 2.541 2.547 2.553 2.559 2.564 1 1 2 2 3 4 4 5 5 0.41 2.570 2.576 2.582 2.588 2.594 2.600 2.606 2.612 2.618 2.624 1 1 2 2 3 4 4 5 5 0.42 2.630 2.636 2.642 2.649 2.655 2.661 2.667 2.673 2.679 2.685 1 1 2 2 3 4 4 5 6 0.43 2.692 2.698 2.704 2.710 2.716 2.723 2.729 2.735 2.742 2.748 1 1 2 3 3 4 4 5 6 0.44 2.754 2.761 2.767 2.773 2.780 2.786 2.793 2.799 2.805 2.812 1 1 2 3 3 4 5 5 6 0.45 2.818 2.825 2.831 2.838 2.844 2.851 2.858 2.864 2.871 2.877 1 1 2 3 3 4 5 5 6 0.46 2.884 2.891 2.897 2.904 2.911 2.917 2.924 2.931 2.938 2.944 1 1 2 3 3 4 5 5 6 0.47 2.951 2.958 2.965 2.972 2.979 2.985 2.992 2.999 3.006 3.013 1 1 2 3 3 4 5 6 6 0.48 3.020 3.027 3.034 3.041 3.048 3.055 3.062 3.069 3.076 3.083 1 1 2 3 4 4 5 6 6 0.49 3.090 3.097 3.105 3.112 3.119 3.126 3.133 3.141 3.148 3.155 1 1 2 3 4 4 5 6 7 0.50 3.162 3.170 3.177 3.184 3.192 3.199 3.206 3.214 3.221 3.228 1 1 2 3 4 4 5 6 7


1647 ANTILOG TABLE: 10P = N Thousandth Parts P 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

0.51 3.236 3.243 3.251 3.258 3.266 3.273 3.281 3.289 3.296 3.304 1 2 2 3 4 5 5 6 7 0.52 3.311 3.319 3.327 3.334 3.342 3.350 3.357 3.365 3.373 3.381 1 2 2 3 4 5 5 6 7 0.53 3.388 3.396 3.404 3.412 3.420 3.428 3.436 3.443 3.451 3.459 1 2 2 3 4 5 6 6 7 0.54 3.467 3.475 3.483 3.491 3.499 3.508 3.516 3.524 3.532 3.540 1 2 2 3 4 5 6 7 7 0.55 3.548 3.556 3.565 3.573 3.581 3.589 3.597 3.606 3.614 3.622 1 2 2 3 4 5 6 7 7 0.56 3.631 3.639 3.648 3.656 3.664 3.673 3.681 3.690 3.698 3.707 1 2 3 3 4 5 6 7 8 0.57 3.715 3.724 3.733 3.741 3.750 3.758 3.767 3.776 3.784 3.793 1 2 3 3 4 5 6 7 8 0.58 3.802 3.811 3.819 3.828 3.837 3.846 3.855 3.864 3.873 3.882 1 2 3 4 4 5 6 7 8 0.59 3.890 3.899 3.908 3.917 3.926 3.936 3.945 3.954 3.963 3.972 1 2 3 4 5 5 6 7 8 0.60 3.981 3.990 3.999 4.009 4.018 4.027 4.036 4.046 4.055 4.064 1 2 3 4 5 6 7 7 8 0.61 4.074 4.083 4.093 4.102 4.111 4.121 4.130 4.140 4.150 4.159 1 2 3 4 5 6 7 8 9 0.62 4.169 4.178 4.188 4.198 4.207 4.217 4.227 4.236 4.246 4.256 1 2 3 4 5 6 7 8 9 0.63 4.266 4.276 4.285 4.295 4.305 4.315 4.325 4.335 4.345 4.355 1 2 3 4 5 6 7 8 9 0.64 4.365 4.375 4.385 4.395 4.406 4.416 4.426 4.436 4.446 4.457 1 2 3 4 5 6 7 8 9 0.65 4.467 4.477 4.487 4.498 4.508 4.519 4.529 4.539 4.550 4.560 1 2 3 4 5 6 7 8 9 0.66 4.571 4.581 4.592 4.603 4.613 4.624 4.634 4.645 4.656 4.667 1 2 3 4 5 6 8 9 100.67 4.677 4.688 4.699 4.710 4.721 4.732 4.742 4.753 4.764 4.775 1 2 3 4 5 7 8 9 100.68 4.786 4.797 4.808 4.819 4.831 4.842 4.853 4.864 4.875 4.887 1 2 3 4 6 7 8 9 100.69 4.898 4.909 4.920 4.932 4.943 4.955 4.966 4.977 4.989 5.000 1 2 3 5 6 7 8 9 100.70 5.012 5.023 5.035 5.047 5.058 5.070 5.082 5.093 5.105 5.117 1 2 4 5 6 7 8 9 110.71 5.129 5.140 5.152 5.164 5.176 5.188 5.200 5.212 5.224 5.236 1 2 4 5 6 7 8 10 110.72 5.248 5.260 5.272 5.284 5.297 5.309 5.321 5.333 5.346 5.358 1 2 4 5 6 7 9 10 110.73 5.370 5.383 5.395 5.408 5.420 5.433 5.445 5.458 5.470 5.483 1 3 4 5 6 8 9 10 110.74 5.495 5.508 5.521 5.534 5.546 5.559 5.572 5.585 5.598 5.610 1 3 4 5 6 8 9 10 120.75 5.623 5.636 5.649 5.662 5.675 5.689 5.702 5.715 5.728 5.741 1 3 4 5 7 8 9 11 120.76 5.754 5.768 5.781 5.794 5.808 5.821 5.834 5.848 5.861 5.875 1 3 4 5 7 8 9 11 120.77 5.888 5.902 5.916 5.929 5.943 5.957 5.970 5.984 5.998 6.012 1 3 4 6 7 8 10 11 120.78 6.026 6.039 6.053 6.067 6.081 6.095 6.109 6.124 6.138 6.152 1 3 4 6 7 8 10 11 130.79 6.166 6.180 6.194 6.209 6.223 6.237 6.252 6.266 6.281 6.295 1 3 4 6 7 9 10 12 130.80 6.310 6.324 6.339 6.353 6.368 6.383 6.397 6.412 6.427 6.442 1 3 4 6 7 9 10 12 130.81 6.457 6.471 6.486 6.501 6.516 6.531 6.546 6.561 6.577 6.592 2 3 5 6 8 9 11 12 140.82 6.607 6.622 6.637 6.653 6.668 6.683 6.699 6.714 6.730 6.745 2 3 5 6 8 9 11 12 140.83 6.761 6.776 6.792 6.808 6.823 6.839 6.855 6.871 6.887 6.902 2 3 5 6 8 10 11 13 140.84 6.918 6.934 6.950 6.966 6.982 6.998 7.015 7.031 7.047 7.063 2 3 5 6 8 10 11 13 150.85 7.079 7.096 7.112 7.129 7.145 7.161 7.178 7.194 7.211 7.228 2 3 5 7 8 10 12 13 150.86 7.244 7.261 7.278 7.295 7.311 7.328 7.345 7.362 7.379 7.396 2 3 5 7 9 10 12 14 150.87 7.413 7.430 7.447 7.464 7.482 7.499 7.516 7.534 7.551 7.568 2 3 5 7 9 10 12 14 160.88 7.586 7.603 7.621 7.638 7.656 7.674 7.691 7.709 7.727 7.745 2 4 5 7 9 11 12 14 160.89 7.762 7.780 7.798 7.816 7.834 7.852 7.870 7.889 7.907 7.925 2 4 5 7 9 11 13 15 160.90 7.943 7.962 7.980 7.998 8.017 8.035 8.054 8.072 8.091 8.110 2 4 6 7 9 11 13 15 170.91 8.128 8.147 8.166 8.185 8.204 8.222 8.241 8.260 8.279 8.299 2 4 6 8 10 11 13 15 170.92 8.318 8.337 8.356 8.375 8.395 8.414 8.433 8.453 8.472 8.492 2 4 6 8 10 12 14 16 180.93 8.511 8.531 8.551 8.570 8.590 8.610 8.630 8.650 8.670 8.690 2 4 6 8 10 12 14 16 180.94 8.710 8.730 8.750 8.770 8.790 8.810 8.831 8.851 8.872 8.892 2 4 6 8 10 12 14 16 180.95 8.913 8.933 8.954 8.974 8.995 9.016 9.036 9.057 9.078 9.099 2 4 6 8 10 13 15 17 190.96 9.120 9.141 9.162 9.183 9.204 9.226 9.247 9.268 9.290 9.311 2 4 6 9 11 13 15 17 190.97 9.333 9.354 9.376 9.397 9.419 9.441 9.462 9.484 9.506 9.528 2 4 7 9 11 13 15 18 200.98 9.550 9.572 9.594 9.616 9.638 9.661 9.683 9.705 9.727 9.750 2 4 7 9 11 13 16 18 200.99 9.772 9.795 9.817 9.840 9.863 9.886 9.908 9.931 9.954 9.977 2 5 7 9 11 14 16 18 21


1648 PART 8: Appendix C – Napier’s Rods 1649 1650

1651 1652 1653

Index Rod


1654


PART 9: Homework 1655 1656 Multiplication with the Gelosia Grid System 1657 1658 Multiply the following using the gelosia grid system. There is a sheet of blank gelosia grids at the 1659 end of the homework section if you need them. 1660 1661 1) 356×706 2) 45,632×673 1662 1663 3) 56.983×2.98 4) 0.03652×0.00045 1664 1665 Show with gelosia grids how to do the following computations. (Recall how the Babylonians did 1666 division.) 1667 1668 5) 784.32÷4 6) 8943.2÷3 1669 1670 7) 873÷20 8) 838.45÷8 1671 1672 1673 Checking Solutions of Cubic Equations 1674 1675 Determine whether or not the given values of x are solutions to the associated equation 1676 1677 9) Is x = 3 a solution of 3 22 5 3 33x x x+ − + = ? 1678 1679 10) Is x = 2 a solution of 3 22 5 3 33x x x+ − + = ? 1680 1681 11) Is x = 4 a solution of 3 22 4 3 1 77x x x− + + = ? 1682 1683 12) Is x = 5 a solution of 3 22 4 3 1 156x x x− + + = ? 1684 1685 1686 Using Cardano’s Formula on Depressed Cubics 1687 1688 Use Cardano’s formula to solve the following equations. Show all steps so that it is clear how 1689 your solution is found. Give your answer in two forms: (1) Exact from with all radicals and 1690 simplified fractions preserved and (2) in decimal form rounded to four decimal places, where 1691 appropriate. 1692 1693 13) 3 4 39x x+ = 14) 3 9 54x x+ = 1694 1695 15) 3 8 32 0x x− + = 16) 3 24 16x x+ = 1696 1697 17) 3 9 12x x= + 18) 3 5 5x x+ = 1698 1699


Using Cardano’s Formula on Non-Depressed Cubics 1700 1701 Solve the following cubic using Cardano’s method of first producing a depressed cubic and then 1702 using Cardano’s equation. Show all algebraic steps. Give your answer in two forms: (1) Exact 1703 from with all radicals and fractions preserved and (2) in decimal form rounded to four decimal 1704 places, where appropriate. 1705 1706 19) 3 23 4 70x x x− + − = 0 1707 1708 20) 3 212 51 64 0x x x+ + − = 1709 1710 21) 3 215 69 135 0x x x− + − = 1711 1712 22) 3 212 56 1176 0x x x− + − = 1713 1714 23) 3 22 24 86 64 0x x x+ + + = 1715 1716 1717 Stevin’s Notation 1718 1719 24) Write in Stevin notation: 678.298 1720 1721 25) Write in Stevin notation: 559.0391 1722 1723 26) Write in Stevin notation: 0.00397 1724 1725 27) What calculation is being demonstrated in the picture shown from 1726

Simon Stevin? Clearly explain your reasoning and conclusions. 1727 1728 28) Use the figure referred to in Problem (27) to show how Stevin might add the following 1729

numbers. Include a final result, written in Stevin’s notation, as part of your work. 1730 1731

34.98 + 238.924 + 183.85 1732 1733

29) Use the figure referred to in Problem (27) to show how Stevin might add the following 1734 numbers. Include a final result, written in Stevin’s notation, as part of your work. 1735

1736 8935.34 + 98.03765 + 3.498 1737

1738 30) Use the figure referred to in Problem (27) to show how Stevin might add the following 1739

numbers. Include a final result, written in Stevin’s notation, as part of your work. 1740 1741

34.982038 + 23.998371 + 33.4981272 1742 1743


Basic Logarithm Rules 1744 1745 Write each of the following as statements with logarithms. 1746 1747 31) 34 64= 32) 1/ 481 3= 1748 1749 33) 5x y= 34) 09 1= 1750 1751

35) 2 1416

− = 36) 1 / 3xA = 1752

1753 Write each of the following as statements with powers. 1754 1755

37) 4log 16 2= 38) 251log 52

= 1756

1757 39) log 3 2z = 40) 100log 1 0= 1758 1759 41) logm x y= 42) log 4 10x = 1760 1761 1762 1763 Use the following rules of logarithms to write the following logarithms as sum, difference, or 1764 product. 1765 1766 1767 1768 1769 1770 1771 1772 43) log(184 332)× 44) 8log72.8 1773 1774

45) 83.5log19.1

46) log 9547 1775

1776

47) 104.8log33.8

48) ( )7log 3.205 1777

1778

49) 3log 129.3 17.04× 50) 8

3log5

1779

1780

log( ) log loglog( / ) log log

log logcb b

AB A BA B A B

n c n

= += −

=


51) Prove the logarithm rule log log logA A BB

= −

. See the proof earlier in the chapter for 1781

log( )AB = logA + logB for an example on how to approach the proof. 1782 1783 Using the Log Tables 1784 1785 Use the log tables to practice finding the values of the following base-10 logarithms. 1786 1787 52) log 7.32 53) log 4.394 1788 1789 54) log 85.66 55) log 79.329 1790 1791 56) log 6345 57) log 9386.55 1792 1793 58) log 18,387.34 59) log 523,764.88 1794 1795 1796 Using the Antilog Tables 1797 1798 Use the antilog tables to practice finding the values of N in the given expressions. 1799 1800 60) log N = 0.536 61) log N = 0.7256 1801 1802 62) log N = 2.349 63) log N = 4.6301 1803 1804 64) log N = 6.8833 65) log N = 7.45838 1805 1806 1807 Using Log and Antilog Tables to Do Calculations 1808 1809 Using the rules of logarithms and the logarithm and antilog tables, compute the values of the 1810 following problems. Show all computations (by hand), including logarithm manipulations. 1811 Express your answer to as many decimal places as the tables yield. Do not use a calculator at 1812 all. 1813 1814 66) 57.3×87.2 67) 732.3×8750 1815 1816 68) 9,838×3,427,200 69) 59×0.0328 1817 1818

70) 48.322.9

71) 5932286.3

1819

1820

72) 3,865,80023971

73) 2.7830.273

1821

1822


74) 34.87 75) 287,450,000 1823 1824 76) 372.6 77) 1/10184.572 1825 1826

78) 3

185.7

79) 62.45 34.8× 1827

1828

80) 10

2

3.87456.1

81) 5490 3599× 1829

1830 1831 Applications of Logarithms 1832 1833 If interest is compounded n times a year, then when you invest P dollars for t years at an interest 1834 rate of r%, the value of your investment after t years is given by the formula: 1835 1836

1ntrA P

n = +

1837

1838 82) Suppose you invest $1000 for one year at a 6% interest rate. Interest is compounded once a 1839

month (12 times a year). 1840 a. Use a calculator to find the value of the account, A, at the end of that year. 1841 b. Without using a calculator, find the value of the account by using log rules and the log 1842

tables. You can do basic computations by hand, but the exponent must be computed with 1843 the log tables. 1844

_____ 1845 1846 83) Suppose you’re lived more than a hundred years ago and you make the interesting 1847

discovery that the time t that it takes for a pendulum to swing from one side to the other 1848

and back (called its period) is given by the formula 232Lt π= . In this equation, L is the 1849

length of the pendulum (feet), π can be approximated as 3.142, and the 32 inside the square 1850 root is the gravity constant for Earth. 1851

a. Suppose you are working on a new, large clock as a gift to a political leader. You will need 1852 an accurate measure of the period of the pendulum you will use for the clock. The 1853 pendulum is 15 feet long. Use log rules, log tables, and pencil and paper to find the period 1854 of the clock to four decimal places. Do not use a calculator for any calculations. Show all 1855 steps clearly and neatly. 1856

b. Use regular algebra to solve for L in terms of t and π. Do not approximate for a value of π. 1857 Explain what this new equation would tell you. 1858

c. Suppose you are building a clock that is to have a period of approximately t = 2.896 1859 seconds (that’s the total time to go both back and forth). Use log rules, log tables, and 1860 pencil and paper to find the length of the pendulum to four decimal places. Do not use a 1861 calculator for any calculations. Show all steps clearly and neatly. 1862


Writing 1863 1864 Write a short essay on the given topic. It should not be more than one page and if you can type it 1865 (double−spaced), I would appreciate it. If you cannot type it, your writing must be legible. 1866 Attention to grammar is important, although it does not have to be perfect grammatically…I just 1867 want to be able to understand it. 1868 1869 84) Was Cardano justified in publishing the solution to the cubic equation? Did he violate his 1870

oath? Explain your position. 1871 1872 85) Research Cardano’s life and give a brief biography of the man that extends the information 1873

given in this chapter or in class. 1874 1875 86) Research the live of Evariste Galois and give a brief biography of the man that extends the 1876

information given in this chapter or in class. 1877 1878 87) Research the history of the slide rule and give a brief account of its invention, evolution, 1879

use, etc. 1880 1881 88) Research the role of geometry in the Renaissance and give a brief description of what you 1882

find. 1883 1884 89) Research the Golden Ratio (Golden Mean) in nature and art and give a description of what 1885

you find. 1886 1887 90) Research the Golden Ratio (Golden Mean) and its current connection to “Sacred 1888

Geometry” and give a description of what you find. 1889 1890 91) Research the Fibonacci numbers; describe them, where they are found, and how they are 1891

used. If possible, try to make a connection between the Fibonacci numbers and the Golden 1892 Ratio (Golden Mean) 1893

1894


PART 10: Endnotes 1895 1 Solution to Check Point A

Your grid should look like this. The answer is 30,004,188

2 Swetz, Learning Activities in the History of Math, page 181 3 Katz, page 260 4 Picture from http://www-groups.dcs.st-andrews.ac.uk/~history/PictDisplay/Tartaglia.html 5 Picture from http://www-groups.dcs.st-andrews.ac.uk/~history/PictDisplay/Cardan.html 6 Dunham, page 140. 7 Dunham, page 141. 8 Dunham, page 142. 9 Dunham, pp. 142-145. 10 Solution to Check Point B

Answer: You should get x = 3. 11 Solution to Check Point C

You should get x ≈ 1.8113. 12 Solution to Check Point D

You should use x = y – 4 to get a depressed cubic of y3 – 44y = 333. This gives y = 9 as a solution which in turn gives x = 5 as the final solution.

13 Katz, page 365 14 Calinger, page 477 15 http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Stevin.html 16 Solution to Check Point E 3,456.28547 = 3,456 2 8 5 4 7 17 Calinger, page 482 18 http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Viete.html 19 Calinger, page 479 20 http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Napier.html 21 Five Fingers to Infinity, page 428 22 Calinger, page 491 23 Solution to Check Point F

4log 16 y= . The value of y is 2 since 42 = 16. 24 Solution to Check Point G

wy z= 25 Sojlution to Check Point H

log(1298) + log(3892) 26 Solution to Check Point I log(z) – log3.5 27 Solution to Check Point J


2 2 / 33 2log log log3

x x x= =

28 Solution to Check Point K 10log 8.55 0.9320= 29 Solution to Check Point M

Did you get 4.5518? 30 Solution to Check Point N

You should get 1.525. N = 100.1834 31 Solution to Check Point O

419,800 32 Solution to Check Point P

You should get a total of 21.2153 from the log tables. The antilog tables gives you 1.642×1021 as a final answer.

33 Solution to Check Point Q You should get a total of 1.1362 from the log tables. The antilog tables gives you 13.689 as a final answer.

34 Solution to Check Point R The log tables should give you 24.4446. The antilog tables gives a final answer of 2.784×1024 which is off by quite a bit.

35 Solution to Check Point S The log tables should give you 2.0426. The antilog tables give a final answer of 110.4

36 http://www.cs.nyu.edu/courses/spring00/V22.0004-002/history/napier2.html 37 CRC Handbook, page 12

2 History of Math for the Liberal Arts CHAPTER 6 · PDF file41 The Slide Rule ... Aryabhata advanced trigonometry, ... 183 represents the ones place, we add up all numbers in that

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