Key Stone Problem… Key Stone Problem… next Set 7 Part 2 © 2007 Herbert I. Gross.

Key Stone Problem…Key Stone Problem…

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Set 7 Part 2© 2007 Herbert I. Gross

You will soon be assigned five problems to test whether you have internalized the

material in Lesson 7 part 2 of our algebra course. The Keystone Illustration below is

a prototype of the problems you'll be doing. Work out the problem on your own.

Afterwards, study the detailed solutions we've provided. In particular, notice that several different ways are presented that

could be used to solve the problem.

Instructions for the Keystone Problem

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© 2007 Herbert I. Gross

As a teacher/trainer, it is important for you to understand and be able to respond

in different ways to the different ways individual students learn. The more ways

you are ready to explain a problem, the better the chances are that the students

will come to understand.

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(a) For what value of n is it

true that … 161/2 = 2n

Keystone Illustration for Lesson 7 Part 2

Answer: n = 2© 2007 Herbert I. Gross

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Solution for Part a:

We saw in the lesson that raising a number to the one-half power means the same thing

as taking the (positive) square root of the number.

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Since the positive square root of 16 is 4, 161/2 = 4

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However, the problem wants the answer in the form 2 n; and since 4 = 22, we may

rewrite the equation in the equivalentform 161/2 = 22

Alternative Solution:Even though there may be only one correct answer. There are usually many paths that

lead to it. For example, one might recognize that 16 = 42, whereupon…

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161/2 = (42)1/2

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= 42 × 1/2

= 41

= 4

= 22

Alternative Solution:

Using the result of (4) we may rewrite…

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2 = n

22 = 2n

4 = 2n161/2 = 2n

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Special Note

• The fact, for example, that 52 = 25 does not mean that if x2 = 25, x then must equal 5. In fact since ( -5)2 is also 25, it means

that if x2 = 25, then x can be either +5 or -5. What this means in terms of our

input/output model is that for certain “recipes” it is possible for two different

inputs to yield the same output.

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Special Note

• In deducing from the equation (4 = 2n) that n = 2, we assumed that there was only

one value of n for which 2n = 4. In thecase of exponents this is true. In other words, if b represents any base, and if

bn = bm , then n must be equal to m.

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Special Note

• While a rigorous proof of this is beyond the scope of this course, an intuitive way

to see that this is plausible is to thinkin terms of compound interest. Namely if

the rate of interest doesn't change the value of the investment continually

increases with time.

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Note

• There is no need to memorize the fact that raising a number to the one-half

power means the same thing as taking the(positive) square root of the number.

Rather we can derive the result using our basic rules for the arithmetic of exponents.

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Note

• For example, starting with the propertybm × bn = bm+n

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we see that if we let b = 16, and m and n both equal = 1/2, the property becomes…

b m × b n =16 161/2 1/2

16 1/2 + 1/2 = 16 1 = 16

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Special Note

• We see from 161/2 × 161/2 = 161 = 16 that 161/2 is that (positive) number which

when multiplied by itself is 16; that is, it is the square root of 16.

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Special Note

• We could also have looked at the property (bm)n = bm×n‚ with b = 16, m =1/2

and n = 2 to obtain…

( b m ) n =16 1/2 2

16 1/2 × 2 = 16 1 = 16

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Note

• In summary, knowing the basic properties and wanting to ensure that we preserve

them allows us to derive the rules ofarithmetic for fractional and/or negative

exponents.• We can generalize the result of what

happens when we have unit fractions (that is, fractions whose numerator is 1) as

exponents. Namely, for any non-zero value of n, 1/n × n = 1.

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Special Note

• So, for example, if we let n = 5, we see that … ( b 1/5 ) n =5

b 1/5 × 5 = b 1 = b

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In other words b1/5 is that number which when raised to the 5th power is equal to b. This number is known as the fifth root of b and is written as √b .

5

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Note

• So, for example, the fact that 25 = 32 means that 2 is the 5 fifth root of 32. One

way to check this is by using the calculator and showing that 321/5 = 2. Since 1/5 = 0.2, we can use the following sequence of key

strokes…

3 2 xy 0.2 =

and we will see that 2 appears in the answer display.

2

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(b) For what value of n is it true that …

161/2 × 2-3 = 2n


Answer: n = -1© 2007 Herbert I. Gross

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Solution for Part b:

From part (a) we already know that

161/2 = 4 = 22

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in the form…

22 × 2-3 = 2n

Hence, we may rewrite the equation…

161/2 × 2-3 = 2n

Solution for Part b:

We may now use our extended rules for the arithmetic of exponents to rewrite 22 × 2-3 as…

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If we now replace 22 × 2-3 by its value, we obtain…

2-1

22 + -322 × 2-3

And we see that… n = -1

= 2n

An Alternative Solution for Part b:

We could also have used the facts that…

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2-3 = 1 ÷ 23 = 1 ÷ 8 = 1/8 ,and 161/2 = 4

to rewrite 161/2 × 2-3 in the form 4 ÷ 8 = 1/2

and then use the fact that…

1/2 = 1 ÷ 2 = 1 ÷ 21 = 2-1

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Note

• There is a tendency for beginning students to think that 2-1 is negative. Keep

in mind that if n is any positive number 2n is positive. Hence, its reciprocal is also positive. Since raising a number to a

negative power means taking the reciprocal of that number it means that 2 raised to any

negative power is a positive number.

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Note

• Notice that 20 = 1. Therefore, if n is less than 0 (that is, if n is negative) 2n

will be less than 1. Moreover as a number increases its reciprocal

decreases.

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Note

• The above discussion is not limited to having 2 as the base. It applies to any positive number b.

The key point is that if n is a very large positive number, b-n is a positive number whose value is close to 0.

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(c) For what value of n is it true that …

161/2 × 2-3 ÷ 2-6 = 2n



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Prelude to Solution for Part c:

In showing why 2n × 2-n = 20 = 1, there is a tendency to think of n as denoting a positive number. However n can be

negative, in which case -n (that is, the opposite of n) is positive. The point is that n + -n = 0 regardless of whether or not n is

positive, and 20 = 1.

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Prelude to Solution for Part c:

More generally if b is any non-zero number and n is any number, then bn and b-n are reciprocals of one another. In particular, dividing a number by b-n means the same

thing as multiplying the number by bn.

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Solution for Part c:

The expression 161/2 × 2-3 ÷ 2-6 contains only the operations of multiplication and

division. Hence, by our PEMDAS agreement we perform the arithmetic from left to right. In other words we may read it

as if it were…

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(

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161/2 × 2-3 ÷ 2-6)


In earlier parts of this problem, we showed that 161/2 = 22. And as explained in the above prelude dividing by 2-6 is equivalent to multiplying by 26.

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(

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22

)

Therefore, we may rewrite the expression in the equivalent form…

× 2-3 × 26(161/2 × 2-3 ) ÷ 2-6

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Since 22 × 2-3 = 22 + -3 = 2-1, we may rewrite the expression as…

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which is equal to…

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2-1+ 6

2-1 × 26(22 × 2-3 ) ÷ 2-6

= 25


In summary,

(161/2 × 2-3 ) ÷ 2-6 = 25

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If we now replace (161/2 × 2-3) ÷ 2-6 = 2n

by its value above, we see that 25 = 2n; so

n = 5.

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An Alternative Solution for Part c:

When in doubt we can always resort to the basic definitions. For example, we already know that 161/2 = 4, 2-3 – 1/8, and 2-6 = 1/2

6 = 1/64.

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Therefore we may rewrite…

(4 × 1/8) ÷ 1/64(161/2 × 2-3 ) ÷ 2-6

An Alternative Solution for Part c:

4 × 1/8 = 1/2 and dividing by 1/64 is the same as multiplying by 64. Therefore, we may rewrite the expression...

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as…

(1/2) × 64(4 × 1/8) ÷ 1/64

If we remember that 32 = 25, we see that n must be 5.

= 32

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Note

• Of course the above approach may be tedious at times, but (1) it always works and (2) by doing things the long” way a

few times may lead to a better internalization of the properties of the

arithmetic of exponents.

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NoteMultiplying Like

Bases

• The fact that multiplication is associative and commutative (meaning that we can

group the factors in any way that we wish) allows us to generalize some of the rules

for exponents.

For example, in demonstrating how to multiply like bases, we used only

examples that involved two factors.

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Bases

• For example, we saw that 23 × 24 = 23+4. However we did not look at an example of

the form 23 × 24 × 25.

Notice that the approach we used in the case of showing that 23 × 24 = 23+4 works in

exactly the same way when there are more factors.

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Bases

• For example, using the definition of exponents we may write 23 × 24 × 25 in the form…

(2 × 2 × 2) × (2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)

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and verify that we have (3 + 4 + 5) factors of 2.

3 4 5

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Bases

• Therefore, in doing Part (c) when we were faced with the expression…

22 × 2-3 ÷ 2-6

We could have rewritten this expression in one step as…

22 + -3 +6 = 25

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(d) For what value of n is it true that …

161/2 × 2-3 ÷ 2-6 + 23 = 2n



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Solution for Part d:

Using PEMDAS, we raise to powers, multiply and divide before we add. Therefore, we may rewrite…

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as …

32 + 8161/2 × 2-3 ÷ 2-6 + 23

We know that 23 = 8, and from part (c) we know that (161/2 × 2-3 ÷ 2-6) = 32. Therefore, we may rewrite the expression as…

= 40( )

In summary, if 161/2 × 2-3 ÷ 2-6 + 23 = 2n, then n = 40.

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Note

• Our rules for exponents apply to multiplication, division, and raising to powers. There are no “nice” rules for

addition and/or subtraction.

• Therefore, in problems such as part (d) it is very important to pay attention to plus and minus signs because they separate

terms.

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Key Stone Problem… Key Stone Problem… next Set 7 Part 2 © 2007 Herbert I. Gross.

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