Slide 4.5- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Slide 4.5- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Exponential and Logarithmic Equations

Learn to solve exponential equations.

Learn to solve applied problems involving exponential equations.

Learn to solve logarithmic equations.

Learn how to use the logistic growth model.

SECTION 4.5

1

2

3

4


EXAMPLE 1Solving an Exponential Equation of “Common Base” Type

a. 25x 125Solve each equation.

b. 9x 3x1

Solution

a. 52 x53

52 x 53

2x 3

x 3

2

b. 32 x3x1

32 x 3x1

2x x 1

2x x 1

x 1Solution set is {1}.Solution set is

3

2

.


EXAMPLE 2Solving Exponential Equations by “Taking Logarithms”

a. 2x 15

Solve each equation and approximate the results to three decimal places.

b. 52x 2 17

Solutiona. 2x 15

ln 2x ln15

x ln 2 ln15

x ln15

ln 23.907


EXAMPLE 2Solving Exponential Equations by “Taking Logarithms”

b. 52x 3 17

2x 3 17

5

ln 2x 3 ln17

5

x 3 ln 2 ln17

5

Solution continued

x 3 ln

175

ln 2

x ln

175

ln 2 3

x 4.766


PROCEDURE FOR SOLVING EXPONENTIAL EQUATIONS

Step 1. Isolate the exponential expression on

one side of the equation.

Step 2. Take the common or natural logarithm

of both sides of the equation in Step 1.

Step 3. Use the power rule log a M r = r log a M

to “bring down the exponent.”

Step 4. Solve for the variable.


EXAMPLE 3Solving an Exponential Equation with Different Bases

ln 52 x 3 ln 3x1

2x 3 ln 5 x 1 ln 3

2x ln 5 3ln 5 x ln 3 ln 3

2x ln 5 x ln 3 ln 3 3ln 5

x 2 ln 5 ln 3 ln 3 3ln 5

x ln 3 3ln 5

2 ln 5 ln 32.795

Solve the equation 52x–3 = 3x+1 and approximate the answer to three decimal places.

When different bases are involved begin with Step 2.

Solution


EXAMPLE 4 An Exponential Equation of Quadratic Form

Solve the equation 3x – 8•3–x = 2.

3x 3x 83 x 2 3x 32 x 830 23x

32 x 8 23x

32 x 23x 8 0

Solution

This equation is quadratic in form. Let y = 3x then y2 = (3x)2 = 32x. Then,

32 x 23x 8 0.



Solution continued

32 x 23x 8 0

y2 2y 8 0

y 2 y 4 0

y 2 0 or y 4 0

y 2 or y 4

3x 2 or 3x 4

But 3x = –2 is not possible because 3x > 0 for all numbers x. So, solve 3x = 4 to find the solution.



Solution continued

3x 4

ln 3x ln 4

x ln 3 ln 4

x ln 4

ln 3x 1.262


EXAMPLE 5 Solving a Population Growth Problem

The following table shows the approximate population and annual growth rate of the United States and Pakistan in 2005.

Country PopulationAnnual

Population Growth Rate

United States 295 million 1.0%

Pakistan 162 million 3.1%



Use the alternate population model P = P0(1 + r)t and the information in the table, and assume that the growth rate for each country stays the same. In this model, P0 is the initial population and t is the time in years since 2005.

a. Use the model to estimate the population of each country in 2015.

b. If the current growth rate continues, in what year will the population of the United States be 350 million?



c. If the current growth rate continues, in what year will the population of Pakistan be the same as the population of the United States?

P P0 1 r t.

Solution

Use the given modela. US population in 2005 is P0 = 295. The

year 2015 is 10 years from 2005.P 295 1 0.01 10 325.86 million

Pakistan in 2005 is P0 = 162.P 162 1 0.31 10 219.84 million



Solution continued

350 295 1 0.01 t

350

295 1.01 t

ln350

295

ln 1.01 t

b. Solve for t to find when the United States population will be 350.

ln350

295

t ln 1.01

t ln

350295

ln 1.01 17.18

Somewhere in the year 2022 (2005 + 17.18) the United States population will be 350.



Solution continued

295 1 0.01 t 162 1 0.031 t

295 1.01 t 162 1.031 t

295

162

1.031

1.01

t

ln295

162

ln

1.031

1.01

t

c. Solve for t to find when the population will be the same in the two countries.



Solution continued

ln295

162

t ln

1.031

1.01

t ln

295162

ln1.0311.01

29.13

Somewhere in the year 2034 (2005 + 29.13) the two populations will be the same.


SOLVING LOGARITHMS EQUATIONS

log2 x 4 means x 24 16

Equations that contain terms of the form log a x are called logarithmic equations.

To solve a logarithmic equation we write it in the equivalent exponential form.

log2 x 4 log3 2x 1 log3 x 2 log2 x 3 log2 x 4 1


EXAMPLE 6 Solving a Logarithmic Equation

Solve: 4 3log2 x 1.

Since the domain of logarithmic functions is positive numbers, we must check our solution.

Solution4 3log2 x 1

3log2 x 1 4 3

log2 x 1

x 2 1

x 1

2


EXAMPLE 6 Solving a Logarithmic Equation

Solution continued

1

2

.The solution set is

1

2.Check x =

4 3log2 x 1

4 3log2

1

21

4 3log2 2 1 1

4 3 1 log2 2 1

4 3 1

1 1

?

?

?

?


EXAMPLE 7Using the One-to-One Property of Logarithms

Solve: log4 x log4 x 1 log4 x 1 log4 6.

Solution

log4 x log4 x 1 log4 x 1 log4 6

log4 x x 1 log4 6 x 1 x x 1 6 x 1

x2 x 6x 6

x2 5x 6 0

x 2 x 3 0



Solution continued

x 2 0 or x 3 0

x 2 or x 3

Check x = 2:log4 x log4 x 1 log4 x 1 log4 6

log4 2 log4 2 1 log4 2 1 log4 6

log4 2 log4 3 log4 1 log4 6

log4 23 0 log4 6

log4 6 log4 6

?

?

?

?



Solution continued

The solution set is {2, 3}.

Check x = 3:

log4 x log4 x 1 log4 x 1 log4 6

log4 3 log4 31 log4 3 1 log4 6

log4 3 log4 4 log4 2 log4 6

log4 34 log4 26 log4 12 log4 12

?

?

?

?


EXAMPLE 8 Using the Product and Quotient Rules

Solve: a. log2 x 3 log2 x 4 1

b. log3 x 1 log3 x 3 1.Solution

a. log2 x 3 log2 x 4 1

log2 x 3 x 4 1

x 3 x 4 21

x2 7x 12 2

x2 7x 10 0

x 2 x 5 0



Solution continued

x 2 0 or x 5 0

x 2 or x 5

Check x = 2:

Logarithms are not defined for negative numbers, so x = 2 is not a solution.

log2 x 3 log2 x 4 1

log2 2 3 log2 2 4 1

log2 1 log2 2 1

?



Solution continued

The solution set is {5}.

Check x = 5:

log2 x 3 log2 x 4 1

log2 5 3 log2 5 4 1

log2 2 log2 1 1

1 0 1

1 1

?

?



Solution continued

log3

x 1

x 3

1

x 1

x 3

31

x 1 3 x 3 x 1 3x 9

8 2x

4 x

b. log3 x 1 log3 x 3 1



Solution continued

The solution set is {4}.

Check x = 4:

log3 x 1 log3 x 3 1

log3 4 1 log3 4 3 1

log3 3 log3 1 1

1 0 1

1 1

?

?

?


EXAMPLE 9 Using the Logistic Growth Model

Suppose the carrying capacity M of the human population on Earth is 35 billion. In 1987, the world population was about 5 billion. Use the logistic growth model of P. F. Verhulst to calculate the average rate, k, of growth of the population, given that the population was about 6 billion in 2003.

P t M

1 ae kt



Solution

We have t = 0 (1987), P(t) = 5 and M = 35.

5 35

1 ae k 0 35

1 a

5 1 a 35

1 a 7

a 6

P t 35

1 6e kt .We now have



Solution continued

6 35

1 6e 16t

6 1 6e 16t 35

6 36e 16t 35

36e 16t 29

Solve for k given t = 16 (for 2003) and P(t) = 6.

e 16t 29

36

16k ln29

36

k 1

16ln

29

36

0.0135

The growth rate was approximately 1.35%.

Documents

Slide 4.5- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley