Solving Equations Involving Logarithmic and Exponential Functions On completion of this module you will be able to: convert logarithmic with bases other.

Solving Equations Involving Logarithmic and Exponential Functions

On completion of this module you will be able to: convert logarithmic with bases other than 10 or e use the inverse property of exponential and logarithmic functions to simplify equations understand the properties of logarithms use the properties of logarithms to simplify equations solve exponential and logarithmic equations

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Bases other than 10 or e

Most calculators have log x (base 10) and ln x (base e). How can we solve equations involving bases other than 10 or e? One way is using the change of base rule:

log ln log or log

log lnb b

x xx x

b b

2

6

5

4

2

1

3

1. log 40

2. log 25

3. log 16

14. log

165. log 9

Example

3

6

log 40 ln 401. log 40 or

log6 ln 6

2.0588 to 4 decimal places

Answer

25

log 252. log 25 =2 note that 5 =25

log5

24

log163. log 16 =2 note that 4 =16

log 4

4

42

1log1 1164. log 4 note that 2 =

16 log 2 16

Answer

1

3

log95. log 9 2

1log

3

5

Using the inverse property

When an exponential function and a logarithmic function have the same base, they are inverses and so effectively cancel each other out.

ExampleSolve for x: log 1.5.x

6

We can’t divide by log! Use the exponential function with the same base (10) – called taking the anti-log. The left and right sides of the equation become exponents with a base of 10

log 1.510 10x

log 1.5x

1.510 31.6228x

7

ExampleSolve for x:AnswerQuick solution is to rearrange using the definition of logs: Alternative:

Now rearrange to isolate the variable:

3log 4.x

43log 4 3 81.x x x

3

loglog 4 4

log3

xx

log 4log3x

8

Take anti-logs:

log 4log3x

log 4log3

4log3

10 10

10 81

x

x

9

ExampleSolve:AnswerWe have an exponential function (base e) which we can cancel out by taking the logarithm with the same base (ln x):

3 2.xe

3 2xe

3ln ln 2xe 3 ln 2x

ln 20.231049 to 6 decimal places

3x

10

log log logmn m n

Rule 1

Properties of logarithms

Example 1

(Since 84 = 12 7)

log84 log12 log7

11

log log logmn m n

Rule 1

Example 2Solve for x:Answer

log log 2 2x x

log log 2 2 log 2 2x x x x

2log 2 2x 2log 2 210 10

x

22 100x 2 50x

7.0711 to 4 decimal placesx 12

Note that although both +7.0711 and −7.0711 square to give 50, only +7.0711 solves the original equation.Check:

as required, but

is undefined.Always check that your answer solves the original problem!!

log 7.0711... log 2 7.0711 2

log 7.0711... log 2 7.0711

log log 2 2x x

13

log log logm

m nn

Rule 2

Example9

log log9 log66

0.9542425 0.7781512

0.1760912

9log log1.5

6

0.1760912

or

14

log logrm r m

Rule 3

Example

log 3 6x

log 3 6x

6

12.5754log 3

x

15

1log log m

m

Rule 4

Example

1log log5 0.69897

5

Note: Rules 1 to 4 have been expressed in base 10, but are equally valid using any base.

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log1 0 and ln1 0 Rule 5

This rule also works for any base e.g.

since

5 12log 1 0 and log 1=0

0 05 1 and 12 1

17

log10 1 and ln 1 e Rule 6

Rule 6 also extends to other bases. Whenever we take the log of the same number as the base, then the answer is 1.e.g. 3 42 10,000log 3 1, log 42 1, log 10,000 1, etc

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log 10 and ln r rr e r

Rule 7

Let’s use Rules 3 and 6 to show why Rule 7 is true.

log 10 log10 1r r r r

ln ln 1re r e r r

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log ln 10 and x xx e x

Rule 8

This uses the concept of log and exponential functions as inverses as we discussed earlier. This rule also works for other bases.

82 loglog2 , 8 , etcxx x x

20

log ln log

log lnb

x xx

b b

Rule 9

Recall that this is the change of base formula used earlier.

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Example

log log 2 1x

log 2 1x

log 2 110 10x

2 10x 5x

22

Example

34 7x

3log 4 log7x

3 log 4 log7x

log73

log 4x

log73 1.5963

log 4x

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Example3 17 2 1xe 3 17 3xe

3 1 3

7xe

3 1 3ln ln

7xe

33 1 ln

7x

33 ln 1

7x

3ln 1

7 0.05093

x

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1. log log logmn m n

2. log log logm

m nn

3. log logrm r m 1

4. log log mm

5. log1 0 and ln1 0 6. log10 1 and ln 1e

7. log 10 and lnr rr e r log ln8. 10 and x xx e x

log ln9. log

log lnb

x xx

b b

Summary: Rules of Logarithms

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In solving equations which involve exponential and logarithmic terms, the following properties allow us to remove such terms and so simplify the equation.

. then, If

. then , log log If

nmbb

nmnmnm

Exponential and logarithmic equations

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Example

log 4 1 log 2x x

4 1 2x x

3 1x 1

3x

27

Example

1255 xx

)1(255 xx

22 xx

2 x

2x

28

Example

It doesn’t matter whether base 10 or base e is used, the result will be the same.

8.5 12x

log 8.5 log12x

log8.5 log12x log12

log8.5

1.0791812

0.92941891.1611354

x

ln 8.5 ln12x

ln8.5 ln12x ln12

ln8.52.4849066

2.14006621.1611354

x

The numbers are different but the result is the same.

Base 10 Base e

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Example

The demand equation for a consumer product is

Solve for p and express your answer in terms of common logarithms. Evaluate p to two decimal places when q = 60.

80 2 .pq

30

pq 280 qp 802

)80log()2log( qp

)80log(2log qp

2log

)80log( qp

Answer

31

2log

)6080log(

,60When

p

q

3219.4

places) decimal 2 (to

32.4

32

Suppose that the daily output of units of a new product on the tth day of a production run is given by:

Such an equation is called a learning equation and indicates as time progresses, output per day will increase.

This may be due to a gain in a worker’s proficiency at his or her job.

Example

0.2500 1 tq e

33

Determine, to the nearest complete unit, the output on (a) the first day and (b) the tenth day after the start of a production run.

(c) After how many days will a daily production run of 400 units be reached? Give your answer to the nearest day.

Example (continued)

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a) On the first day or production, t =1, so the daily output will be

b) When t =10,

Note that since the answers to parts (a) and (b) are the number of units of a new product, we have rounded these to the nearest whole unit.

Answer 0.2500 1 tq e

0.2 1500 1 90.6346 91 units.q e

0.2 10500 1 432.3324 432 units.q e

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c) The production run will reach 400 units when q = 400 or at

0.2500 1 tq e

0.2400 500 1 te 0.2400

1500

te

0.2 4 11

5 5te

0.2 1ln ln

5te

36

10.2 ln

5t

1ln

5 8.0472 9 days0.2

t

37

Notice that the question requires the answer to be rounded to the nearest whole day.

If the answer were round to 8 days,

so production has not quite reached 400.

For this reason we round the answer to 9 days, even though production will be well passed 400 by the end of the 9th day.

0.2 8500 1 399.0517 unitsq e

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As always, we must check that the mathematically obtained solution answers the original question.

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