12.6 Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following:

Solve Logarithmic Equations

In Section 12.4 we solved logarithmic equations by changing a logarithmic equation to an exponential equation. That is, we used the definition of a logarithm:

y = log a x is equivalent to x = a ^y a 7 0, a  1

For example, to solve the equation log 2 11 - 2x2 = 3, we use the equivalent exponential equation 1 - 2x = 2 ³ and solve for x.

log 2 11 - 2x2 = 3 1 - 2x = 2 ³

-2x = 7 x = - 7

2

You should check this solution for yourself.

For most logarithmic equations, some manipulation of the equation (usually using properties of logarithms) is required to obtain a solution. Also, to avoid extraneous solutions with logarithmic equations, we determine the domain of the variable first.

Our practice will be to solve equations, whenever possible, by finding exact solutions using algebraic methods and exact or approximate solutions using a graphing utility. When algebraic methods cannot be used, approximate solutions will be obtained using a graphing utility. The reader is encouraged to pay particular attention to the form of equations for which exact solutions are possible.

We begin with an example of a logarithmic equation that requires using the fact that a logarithmic function is a one-to-one function.

If log a M = log a N, then M = N M, N, and a are positive and a  1

Solving a l ogarithmic equation Solve: 2 log 5 x = log 5 9

1

Change to an exponential expression.

Simplify.

Divide both sides by 22.

e x a mp l e 1

Now Work the ‘Are You Prepared?’ problems on page 816.

OBJECTIVES 1 Solve Logarithmic Equations (p. 811) 2 Solve Exponential Equations (p. 813)

3 Solve Logarithmic and Exponential Equations Using a Graphing Utility (p. 815)

• Solve Quadratic Equations (Section 7.2, pp. 476–483) • Solve Equations Quadratic in Form (Section 7.4, pp. 497–499)

PREPARING FOR THIS SECTION Before getting started, review the following:

12.6 Logarithmic and Exponential Equations

Now Work p r o b l e m 1 3 r log a M = log a M ^r

a lgebraic Solution

Note that the domain of the variable in this equation is x 7 0.

Because each logarithm is to the same base, 5, we can obtain an exact solution as follows:

2 log 5 x = log 5 9 log 5 x ² = log 5 9

x ² = 9 x = 3 or x = -3

Check: 2 log 5 3  log 5 9 log 5 3 ²  log 5 9 log 5 9 = log 5 9 The solution set is {3}.

4

4

4 6 Figure 47 log _a M ^r = r log a M

If log _a M = log a N, then M = N.

Recall that the domain of the variable is x 7 0.

Therefore, -3 is extraneous and we discard it.

Graphing Solution

To solve the equation using a graphing utility, graph Y ₁ = 2 log 5 x = 2 log x

log 5 and Y 2 = log 5 9 = log 9 log 5 , and determine the point of intersection. See Figure 47. The point of intersection is (3, 1.3652124);

so x = 3 is the only solution. The solution set is {3}.

WARNING A negative solution is not automatically extraneous. You must determine whether the potential solution causes the argument of any logarithmic expression in the equation to be negative. ^j

Figure 48

3 3

3 3

a lgebraic Solution

The domain of the variable requires that x + 6 7 0 and x + 2 7 0, so x 7 -6 and x 7 -2. This means any solution must satisfy x 7 -2. To obtain an exact solution, express the left side as a single logarithm. Then change the equation to exponential form.

log ₅ 1x + 62 + log 5 1x + 22 = 1 log ₅ 3 1x + 62 1x + 22 4 = 1

1x + 62 1x + 22 = 5 ¹ = 5 x ² + 8x + 12 = 5

x ² + 8x + 7 = 0

1x + 72 1x + 12 = 0 x = -7 or x = -1

Only x = -1 satisfies the restriction that x 7 -2, so x = -7 is extraneous. The solution set is {-1}, which you should check.

Graphing Solution Graph

log1x + 22

log 5 and Y 2 = 1 and determine the point(s) of intersection. See Figure 48. The point of intersection is

1 -1, 12, so x = -1 is the only solution. The solution set is { -1}.

Y ₁ = log 5 1x + 62 + log 5 1x + 22 = log 1x + 62 log 5 +

log a M + log a N = log a 1MN2 Change to an exponential equation.

Simplify.

Place the quadratic equation in standard form.

Factor.

Zero-Product Property

Solving a l ogarithmic equation Solve: log 5 1x + 62 + log 5 1x + 22 = 1 e x a mp l e 2

Now Work p r o b l e m 2 1

Solving a l ogarithmic equation Solve: ln x + ln1x - 42 = ln1x + 62 e x a mp l e 3

WARNING In using properties of logarithms to solve logarithmic equations, avoid using the property log a x ^r = r log a x, when r is even. The reason can be seen in this example:

Figure 49

7 10

3 5

a lgebraic Solution

The domain of the variable requires that x 7 0, x - 4 7 0, and x + 6 7 0.

As a result, the domain of the variable is x 7 4. Begin the solution using the log of a product property.

ln x + ln1x - 42 = ln1x + 62 ln [x 1x - 42] = ln1x + 62

x 1x - 42 = x + 6 x ² - 4x = x + 6 x ² - 5x - 6 = 0 1x - 62 1x + 12 = 0 x = 6 or x = -1

Since the domain of the variable is x 7 4, discard -1 as extraneous. The solution set is {6}, which you should check.

Graphing Solution

Graph Y ₁ = ln x + ln 1x - 42 and Y 2 = ln 1x + 62 and determine the point(s) of intersection. See Figure 49. The x-coordinate of the point of intersection is 6, so the solution set is {6}.

In M + ln N = ln1MN2 If ln M = ln N, then M = N.

Simplify.

Place the quadratic equation in standard form.

Factor.

Zero-Product Property

Both –9 and 9 are solutions of log 3 x ² = 4 (as you can verify). The process in part (b) does not find the solution –9 because the domain of the variable was further restricted to x 7 0 due to the

application of the property log a x ^r = r log a x. ^j

Now Work p r o b l e m 3 1

Solve Exponential Equations

In Sections 12.3 and 12.4, we solved exponential equations algebraically by expressing each side of the equation using the same base. That is, we used the one- to-one property of the exponential function:

If a ^u = a ^v , then u = v a 7 0, a  1

For example, to solve the exponential equation 4 ^{2x +1} = 16, notice that 16 = 4 ² and apply the property above to obtain 2x + 1 = 2, from which we find x = 1

2 .

For most exponential equations, we cannot express each side of the equation using the same base. In such cases, algebraic techniques can sometimes be used to obtain exact solutions. When algebraic techniques cannot be used, we use a graphing utility to obtain approximate solutions. You should pay particular attention to the form of equations for which exact solutions are obtained.

Solving an exponential equation Solve: 2 ^x = 5

2

e x a mp l e 4

Solve: log ₃ x ² = 4

Solution: The domain of the variable x is all real numbers except 0.

Change to exponential form. (b) log ₃ x ² = 4 Domain of variable is x > 0.

2 log 3 x = 4 log ₃ x = 2 x = 9 (a) log ₃ x ² = 4

x ² = 3 ⁴ = 81 x = -9 or x = 9

log _a x ^r = r log _a x

Now Work p r o b l e m 3 5

Solving an exponential equation Solve: 8 # ^{3 x} = 5

e x a mp l e 5

Figure 50 16

4

4

1

a lgebraic Solution

Since 5 cannot be written as an integer power of 2, write the exponential equation as the equivalent logarithmic equation.

2 ^x = 5

x = log 2 5 = ln 5 ln 2

Change@of@Base Formula 110), Section 6.5

Alternatively, we can solve the equation 2 ^x = 5 by taking the natural logarithm (or common logarithm) of each side. Taking the natural logarithm,

2 ^x = 5 ln 2 ^x = ln 5 x ln 2 = ln 5 x = ln 5 ln 2

 2.322 The solution set is e ln 5

ln 2 f . c

Graphing Solution

Graph Y 1 = 2 ^x and Y 2 = 5 and determine the x- coordinate of the point of intersection. See Figure 50.

The approximate solution, rounded to three decimal places, is 2.322.

If M = N, then ln M = ln N.

In M ^r = r ln M Exact solution

Approximate solution

a lgebraic Solution

Isolate the exponential expression and then rewrite the statement as an equivalent logarithm.

8 # ^{3 x} = 5 3 ^x = 5 8

x = log ₃ a 5 8 b =

ln a 5 8 b ln 3

 -0.428

The solution set is e log 3 a 5 8 b f .

Graphing Solution

Graph Y 1 = 8 # 3 ^x and Y 2 = 5 and determine the x-coordinate of the point of intersection. See Figure 51.

The approximate solution, rounded to three decimal places, is –0.428.

Solve for 3 ^x .

Exact solution

Approximate solution

24

8

2 1

Figure 51

Solving an exponential equation Solve: 5 ^{x -2} = 3 ^{3x +2}

e x a mp l e 6

a lgebraic Solution

Because the bases are different, we first apply property (7), Section 12.5 (take the natural logarithm of each side), and then use appropriate properties of logarithms. The result is a linear equation in x that we can solve.

Graphing Solution

Graph Y 1 = 5 ^{x -2} and Y 2 = 3 ^{3x +2} and determine the

x-coordinate of the point of intersection. See

Figure 52 on the next page.

Now Work p r o b l e m 4 1

The next example deals with an exponential equation that is quadratic in form.

Solving an exponential equation t hat is Quadratic in form Solve: 4 ^x - 2 ^x - 12 = 0

e x a mp l e 7

Figure 53 ₁₀₀

30

4

1

0.0005

4 0

5 = 3 ln 5 ^{x -2} = ln 3 ^{3x +2} 1x - 22 ln 5 = 13x + 22 ln 3 1ln 52x - 2 ln 5 = 13 ln 32x + 2 ln 3 1ln 52x - 13 ln 32x = 2 ln 3 + 2 ln 5

1ln 5 - 3 ln 32x = 21ln 3 + ln 52 x = 2 1ln 3 + ln 52 ln 5 - 3 ln 3

 -3.212

If M = N, ln M = ln N.

ln M ^r = r ln M

Distribute. The equation is now linear in x.

Place terms involving x on the left.

Factor.

Exact solution Approximate solution

The approximate solution, rounded to three decimal places, is -3.212. Note that the y-coordinate, 2.2763E–4, is in scientific notation and means 2.2763 * 10 ^-4 = 0.0002763.

a lgebraic Solution

We note that 4 ^x = 12 ² 2 ^x = 2 ^2x = 12 ^x 2 ² , so the equation is actually quadratic in form, and we can rewrite it as

12 ^x 2 ² - 2 ^x - 12 = 0 ^{Let u} = 2 ^x ; then u ² - u - 12 = 0.

Now we can factor as usual.

12 ^x - 42 12 ^x + 32 = 0 2 ^x - 4 = 0 or 2 ^x + 3 = 0 2 ^x = 4 2 ^x = -3

The equation on the left has the solution x = 2, since 2 ^x = 4 = 2 ² ; the equation on the right has no solution, since 2 ^x 7 0 for all x.

The only solution is 2. The solution set is {2}.

Graphing Solution

Graph Y 1 = 4 ^x - 2 ^x - 12 and determine the x-intercept. See Figure 53. The x-intercept is 2, so the solution set is {2}.

1u - 42 1u + 32 = 0 u - 4 = 0 or u + 3 = 0

u = 2 ^x = 4 or u = 2 ^x = -3

Now Work p r o b l e m 5 3

Solve Logarithmic and Exponential Equations Using a Graphing Utility

The algebraic techniques introduced in this section to obtain exact solutions apply only to certain types of logarithmic and exponential equations. Solutions for other types are usually studied in calculus, using numerical methods. For such types, we can use a graphing utility to approximate the solution.

Solving equations Using a Gra phing Utility Solve: x + e ^x = 2

Express the solution(s) rounded to two decimal places.

The solution is found by graphing Y 1 = x + e ^x and Y 2 = 2. Since Y 1 is an increasing function (do you know why?), there is only one point of intersection for Y 1 and Y 2 .

3

e x a mp l e 8

Solution

Figure 54 shows the graphs of Y 1 and Y 2 . Using the INTERSECT command, the solution is 0.44 rounded to two decimal places.

Now Work p r o b l e m 6 3 Figure 54

4

0

Y ₁  x  e ^x Y ₂  2

0 1

12.6 a ssess Your Understanding

’Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

Skill Building

In Problems 5–32, solve each logarithmic equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places.Verify your results using a graphing utility.

5. log ₄ x = 2 6. log 1x + 62 = 1 7. log ₂ 15x2 = 4

8. log ₃ 13x - 12 = 2 9. log ₄ 1x + 22 = log 4 8 10. log ₅ 12x + 32 = log 5 3 11. 1

2 log ₃ x = 2 log 3 2 12. -2 log 4 x = log 4 9 13. 3 log ₂ x = -log 2 27 14. 2 log ₅ x = 3 log 5 4 15. 3 log ₂ 1x - 12 + log 2 4 = 5 16. 2 log ₃ 1x + 42 - log 3 9 = 2 17. log x + log1x + 152 = 2 18. log x + log 1x - 212 = 2 19. log(2x + 1) = 1 + log(x - 2) 20. log12x2 - log1x - 32 = 1 21. log ₂ 1x + 72 + log 2 1x + 82 = 1 22. log ₆ (x + 4) + log 6 (x + 3) = 1 23. log ₈ (x + 6) = 1 - log 8 (x + 4) 24. log ₅ 1x + 32 = 1 - log 5 1x - 12 25. ln x + ln1x + 22 = 4

26. ln1x + 12 - ln x = 2 27. log ₃ 1x + 12 + log 3 1x + 42 = 2 28. log ₂ 1x + 12 + log 2 1x + 72 = 3 29. log _{1 >3} 1x ² + x2 - log 1 >3 1x ² - x2 = -1 30. log ₄ 1x ² - 92 - log 4 1x + 32 = 3

31. log _a 1x - 12 - log a 1x + 62 = log a 1x - 22 - log a 1x + 32 32. log _a x + log a 1x - 22 = log a 1x + 42

In Problems 33–60, solve each exponential equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places. Verify your results using a graphing utility.

33. 2 ^x-5 = 8 34. 5 ^-x = 25 35. 2 ^x = 10 36. 3 ^x = 14

37. 8 ^-x = 1.2 38. 2 ^-x = 1.5 39. 512 ^3x 2 = 8 40. 0.314 ^0.2x 2 = 0.2

41. 3 ^{1 -2x} = 4 ^x 42. 2 ^{x +1} = 5 ^{1 -2x} 43. a 3

5 b

x = 7 ^{1 -x} 44. a 4

3 b

1-x = 5 ^x

45. 1.2 ^x = 10.52 ^-x 46. 0.3 ^{1 +x} = 1.7 ^{2x -1} 47. p ^{1 -x} = e ^x 48. e ^x+3 = p ^x

49. 2 ^2x + 2 ^x - 12 = 0 50. 3 ^2x + 3 ^x - 2 = 0 51. 3 ^2x + 3 ^x+1 - 4 = 0 52. 2 ^2x + 2 ^x+2 - 12 = 0 53. 16 ^x + 4 ^x+1 - 3 = 0 54. 9 ^x - 3 ^x+1 + 1 = 0 55. 25 ^x - 8 # ^{5 x} = -16 56. 36 ^x - 6 # ^{6 x} = -9 57. 3 # ^{4 x} + 4 # ^{2 x} + 8 = 0 58. 2 # ^{49 x} + 11 # ^{7 x} + 5 = 0 59. 4 ^x - 10 # ^{4 -x} = 3 60. 3 ^x - 14 # ^{3 -x} = 5

In Problems 61–74, use a graphing utility to solve each equation. Express your answer rounded to two decimal places.

61. log ₅ 1x + 12 - log 4 1x - 22 = 1 62. log ₂ 1x - 12 - log 6 1x + 22 = 2

63. e ^x = -x 64. e ^2x = x + 2 65. e ^x = x ² 66. e ^x = x ³

67. ln x = -x 68. ln12x2 = -x + 2 69. ln x = x ³ - 1 70. ln x = -x ²

71. e ^x + ln x = 4 72. e ^x - ln x = 4 73. e ^-x = ln x 74. e ^-x = -ln x

1. Solve x ² - 7x - 30 = 0. (pp. 476–483)

2. Solve 1x + 32 ² - 41x + 32 + 3 = 0. (pp. 497–498)

3. Approximate the solution(s) to x ³ = x ² - 5 using a graphing utility.

4. Approximate the solution(s) to x ³ - 2x + 2 = 0 using a

graphing utility.

In Problems 75–86, solve each equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places.

75. log 2 1x + 12 - log 4 x = 1 76. log 2 13x + 22 - log 4 x = 3 77. log 16 x + log 4 x + log 2 x = 7

78. log ₉ x + 3 log 3 x = 14 79. 1 1 ³ 2 2 ^2-x = 2 ^{x 2} 80. log ₂ x ^{log 2 x} = 4 81. e ^x + e ^-x

2 = 1 82. e ^x + e ^-x

2 = 3 83. e ^x - e ^-x

2 = 2

84. e ^x - e ^-x

2 = -2 85. log ₅ x + log 3 x = 1 86. log ₂ x + log 6 x = 3

[Hint: Change log ₄ x to base 2.]

[Hint: Multiply each side by e ^x .]

[Hint: Use the Change-of-Base Formula.]

87. f1x2 = log 2 1x + 32 and g1x2 = log 2 13x + 12.

(a) Solve f 1x2 = 3. What point is on the graph of f ? (b) Solve g 1x2 = 4. What point is on the graph of g?

(c) Solve f 1x2 = g1x2. Do the graphs of f and g intersect?

If so, where?

(d) Solve 1f + g2 1x2 = 7.

(e) Solve 1f - g2 1x2 = 2.

88. f1x2 = log 3 1x + 52 and g1x2 = log 3 1x - 12.

(a) Solve f 1x2 = 2. What point is on the graph of f ?

(b) Solve g 1x2 = 3. What point is on the graph of g?

(c) Solve f 1x2 = g1x2. Do the graphs of f and g intersect?

If so, where?

(d) Solve 1f + g2 1x2 = 3.

(e) Solve 1f - g2 1x2 = 2.

89. (a) Graph f1x2 = 3 ^{x +1} and g(x) = 2 ^{x +2} , on the same Cartesian plane.

(b) Find the point(s) of intersection of the graphs of f and g by solving f 1x2 = g1x2. Round answers to three decimal places. Label any intersection points on the graph drawn in part (a).

(c) Based on the graph, solve f 1x2 7 g1x2.

90. (a) Graph f1x2 = 5 ^{x -1} and g(x) = 2 ^{x +1} , on the same Cartesian plane.

(b) Find the point(s) of intersection of the graphs of f and g by solving f 1x2 = g1x2. Label any intersection points on the graph drawn in part (a).

(c) Based on the graph, solve f 1x2 7 g1x2.

91. (a) Graph f1x2 = 3 ^x and g(x) = 10 on the same Cartesian plane.

(b) Shade the region bounded by the y-axis, f 1x2 = 3 ^x , and g(x) = 10 on the graph drawn in part (a).

(c) Solve f 1x2 = g1x2 and label the point of intersection on the graph drawn in part (a).

92. (a) Graph f1x2 = 2 ^x and g(x) = 12 on the same Cartesian plane.

(b) Shade the region bounded by the y-axis, f 1x2 = 2 ^x , and g(x) = 12 on the graph drawn in part (a).

(c) Solve f 1x2 = g1x2 and label the point of intersection on the graph drawn in part (a).

93. (a) Graph f1x2 = 2 ^{x +1} and g(x) = 2 ^-x+2 on the same Cartesian plane.

(b) Shade the region bounded by the y-axis, f 1x2 = 2 ^{x +1} , and g(x) = 2 ^-x+2 on the graph draw in part (a).

(c) Solve f 1x2 = g1x2 and label the point of intersection on the graph drawn in part (a).

94. (a) Graph f1x2 = 3 ^-x+1 and g 1x2 = 3 ^{x -2} on the same Cartesian plane.

(b) Shade the region bounded by the y-axis, f 1x2 = 3 ^-x+1 , and g 1x2 = 3 ^{x -2} on the graph draw in part (a).

(c) Solve f 1x2 = g1x2 and label the point of intersection on the graph drawn in part (a).

95. (a) Graph f1x2 = 2 ^x - 4.

(b) Find the zero of f.

(c) Based on the graph, solve f 1x2 6 0.

96. (a) Graph g1x2 = 3 ^x - 9.

(b) Find the zero of g.

(c) Based on the graph, solve g 1x2 7 0.

Applications and Extensions

97. A Population Model The resident population of the United States in 2010 was 309 million people and was growing at a rate of 0.9% per year. Assuming that this growth rate continues, the model P 1t) = 30911.009) ^{t - 2010} represents the population P (in millions of people) in year t.

(a) According to this model, when will the population of the United States be 419 million people?

(b) According to this model, when will the population of the United States be 488 million people?

Source: U.S. Census Bureau

98. A Population Model The population of the world in 2011 was 6.91 billion people and was growing at a rate of 1.14%

per year. Assuming that this growth rate continues, the model P 1t2 = 6.9111.01142 ^t-2011 represents the population P (in billions of people) in year t.

(a) According to this model, when will the population of the world be 9.6 billion people?

(b) According to this model, when will the population of the world be 12 billion people?

Source: U.S. Census Bureau

99. Depreciation The value V of a Chevy Cobalt that is t years old can be modeled by V1t2 = 16,50010.822 ^t .

(a) According to the model, when will the car be worth

$9000?

(b) According to the model, when will the car be worth

$4000?

(c) According to the model, when will the car be worth

$2000?

Source: Kelley Blue Book

Explaining Concepts: Discussion and Writing

101. Fill in reasons for each step in the following two solutions.

Solve: log ₃ 1x - 12 ² = 2

Solution A Solution B

log 3 1x - 1) ² = 2 log 3 1x - 12 ² = 2

1x - 12 ² = 3 ² = 9 _____ 2 log 3 1x - 12 = 2 _____

1x - 12 = { 3 _____ log 3 1x - 12 = 1 _____

x - 1 = -3 or x - 1 = 3 _____ x - 1 = 3 ¹ = 3 _____

x = -2 or x = 4 _____ x = 4 _____

Both solutions given in Solution A check. Explain what caused the solution x = -2 to be lost in Solution B.

‘Are You Prepared?’ Answers

1. {-3, 10} 2. {-2, 0} 3. {-1.43} 4. {-1.77}

100. Depreciation The value V of a Honda Civic DX that is t years old can be modeled by V 1t2 = 16,77510.9052 ^t . (a) According to the model, when will the car be worth

$15,000?

(b) According to the model, when will the car be worth

$8000?

(c) According to the model, when will the car be worth

$4000?

Source: Kelley Blue Book

Determine the Future Value of a Lump Sum of Money

Interest is money paid for the use of money. The total amount borrowed (whether by an individual from a bank in the form of a loan or by a bank from an individual in the form of a savings account) is called the principal. The rate of interest, expressed

1

OBJECTIVES 1 Determine the Future Value of a Lump Sum of Money (p. 818) 2 Calculate Effective Rates of Return (p. 822)

3 Determine the Present Value of a Lump Sum of Money (p. 823) 4 Determine the Rate of Interest or Time Required to Double a Lump

Sum of Money (p. 824)

12.7 Financial Models