Figure 1. Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the “rabbit plague.” (credit: Richard Taylor, Flickr)

In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.

Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.

Use like bases to solve exponential equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b, S, and T, where [latex]b>0,\text{ }b\ne 1[/latex], [latex]{b}^{S}={b}^{T}[/latex] if and only if S = T.

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.

For example, consider the equation [latex]{3}^{4x - 7}=\frac{{3}^{2x}}{3}[/latex]. To solve for x, we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x:

Rewriting Equations So All Powers Have the Same Base

Example 4: Solving an Equation with Positive and Negative Powers

Solve [latex]{3}^{x+1}=-2[/latex].

Show Solution

This equation has no solution. There is no real value of x that will make the equation a true statement because any power of a positive number is positive.

Analysis of the Solution

The figure below shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

Figure 1

 Use logarithms to solve exponential equations

 Use the definition of a logarithm to solve logarithmic equations

We have already seen that every logarithmic equation [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex] is equivalent to the exponential equation [latex]{b}^{y}=x[/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation [latex]{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x - 5\right)=3[/latex]. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x:

Example 12: Using a Graph to Understand the Solution to a Logarithmic Equation

Solve [latex]\mathrm{ln}x=3[/latex].

Show Solution

[latex]\begin{align}\mathrm{ln}x&=3 \\ x&={e}^{3}&& \text{Use the definition of the natural logarithm.} \end{align}[/latex]

Figure 2 represents the graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to 20. In other words [latex]{e}^{3}\approx 20[/latex]. A calculator gives a better approximation: [latex]{e}^{3}\approx 20.0855[/latex].

 

 Use the one-to-one property of logarithms to solve logarithmic equations

Key Equations

Section 5.6 Homework Exercises

1. How can an exponential equation be solved?

2. When does an extraneous solution occur? How can an extraneous solution be recognized?

3. When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

For the following exercises, use like bases to solve the exponential equation.

4. [latex]{4}^{-3v - 2}={4}^{-v}[/latex]

5. [latex]64\cdot {4}^{3x}=16[/latex]

6. [latex]{3}^{2x+1}\cdot {3}^{x}=243[/latex]

7. [latex]{2}^{-3n}\cdot \frac{1}{4}={2}^{n+2}[/latex]

8. [latex]625\cdot {5}^{3x+3}=125[/latex]

9. [latex]\frac{{36}^{3b}}{{36}^{2b}}={216}^{2-b}[/latex]

10. [latex]{\left(\frac{1}{64}\right)}^{3n}\cdot 8={2}^{6}[/latex]

For the following exercises, use logarithms to solve.

11. [latex]{9}^{x - 10}=1[/latex]

12. [latex]2{e}^{6x}=13[/latex]

13. [latex]{e}^{r+10}-10=-42[/latex]

14. [latex]2\cdot {10}^{9a}=29[/latex]

15. [latex]-8\cdot {10}^{p+7}-7=-24[/latex]

16. [latex]7{e}^{3n - 5}+5=-89[/latex]

17. [latex]{e}^{-3k}+6=44[/latex]

18. [latex]-5{e}^{9x - 8}-8=-62[/latex]

19. [latex]-6{e}^{9x+8}+2=-74[/latex]

20. [latex]{2}^{x+1}={5}^{2x - 1}[/latex]

21. [latex]{e}^{2x}-{e}^{x}-132=0[/latex]

22. [latex]7{e}^{8x+8}-5=-95[/latex]

23. [latex]10{e}^{8x+3}+2=8[/latex]

24. [latex]4{e}^{3x+3}-7=53[/latex]

25. [latex]8{e}^{-5x - 2}-4=-90[/latex]

26. [latex]{3}^{2x+1}={7}^{x - 2}[/latex]

27. [latex]{e}^{2x}-{e}^{x}-6=0[/latex]

28. [latex]3{e}^{3 - 3x}+6=-31[/latex]

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

29. [latex]\mathrm{log}\left(\frac{1}{100}\right)=-2[/latex]

30. [latex]{\mathrm{log}}_{324}\left(18\right)=\frac{1}{2}[/latex]

For the following exercises, use the definition of a logarithm to solve the equation.

31. [latex]5{\mathrm{log}}_{7}n=10[/latex]

32. [latex]-8{\mathrm{log}}_{9}x=16[/latex]

33. [latex]4+{\mathrm{log}}_{2}\left(9k\right)=2[/latex]

34. [latex]2\mathrm{log}\left(8n+4\right)+6=10[/latex]

35. [latex]10 - 4\mathrm{ln}\left(9 - 8x\right)=6[/latex]

For the following exercises, use the product or quotient properties of logarithms to solve.

36. [latex]\mathrm{log}(x)+\mathrm{log}(x+9)=1[/latex]

37. [latex]\mathrm{log}(x)+\mathrm{log}(x+3)=1[/latex]

38. [latex]{\mathrm{log}}_{2}(x+5)=3-{\mathrm{log}}_{2}(x+7)[/latex]

39. [latex]{\mathrm{log}}_{3}(x+12)=3-{\mathrm{log}}_{3}(x+6)[/latex]

40. [latex]{\mathrm{log}}_{6}(x+26)-{\mathrm{log}}_{6}(x-9)=2[/latex]

41. [latex]{\mathrm{log}}_{4}(x+12)-{\mathrm{log}}_{4}(x-3)=2[/latex]

42. [latex]\mathrm{log}(4x+5)=1+\mathrm{log}(x-7)[/latex]

43. [latex]\mathrm{log}(8x+1)=1+\mathrm{log}(x-1)[/latex]

44. [latex]{\mathrm{log}}_{6}(x-7)+{\mathrm{log}}_{6}(x-2)-{\mathrm{log}}_{6}(x)=2[/latex]

45. [latex]{\mathrm{log}}_{2}(x-6)+{\mathrm{log}}_{2}(x-4)-{\mathrm{log}}_{2}(x)=2[/latex]

For the following exercises, use the one-to-one property of logarithms to solve.

46. [latex]\mathrm{ln}\left(10 - 3x\right)=\mathrm{ln}\left(-4x\right)[/latex]

47. [latex]{\mathrm{log}}_{13}\left(5n - 2\right)={\mathrm{log}}_{13}\left(8 - 5n\right)[/latex]

48. [latex]\mathrm{log}\left(x+3\right)-\mathrm{log}\left(x\right)=\mathrm{log}\left(74\right)[/latex]

49. [latex]\mathrm{ln}\left(-3x\right)=\mathrm{ln}\left({x}^{2}-6x\right)[/latex]

50. [latex]{\mathrm{log}}_{4}\left(6-m\right)={\mathrm{log}}_{4}3m[/latex]

51. [latex]\mathrm{ln}\left(x - 2\right)-\mathrm{ln}\left(x\right)=\mathrm{ln}\left(54\right)[/latex]

52. [latex]{\mathrm{log}}_{9}\left(2{n}^{2}-14n\right)={\mathrm{log}}_{9}\left(-45+{n}^{2}\right)[/latex]

53. [latex]\mathrm{ln}\left({x}^{2}-10\right)+\mathrm{ln}\left(9\right)=\mathrm{ln}\left(10\right)[/latex]

For the following exercises, solve each equation for x.

54. [latex]\mathrm{log}\left(x+12\right)=\mathrm{log}\left(x\right)+\mathrm{log}\left(12\right)[/latex]

55. [latex]\mathrm{ln}\left(x\right)+\mathrm{ln}\left(x - 3\right)=\mathrm{ln}\left(7x\right)[/latex]

56. [latex]{\mathrm{log}}_{2}\left(7x+6\right)=3[/latex]

57. [latex]\mathrm{ln}\left(7\right)+\mathrm{ln}\left(2 - 4{x}^{2}\right)=\mathrm{ln}\left(14\right)[/latex]

58. [latex]{\mathrm{log}}_{8}\left(x+6\right)-{\mathrm{log}}_{8}\left(x\right)={\mathrm{log}}_{8}\left(58\right)[/latex]

59. [latex]\mathrm{ln}\left(3\right)-\mathrm{ln}\left(3 - 3x\right)=\mathrm{ln}\left(4\right)[/latex]

60. [latex]\mathrm{log_{3}}\left(3x\right)-\mathrm{log_{3}}\left(6\right)=\mathrm{log_{3}}\left(77\right)[/latex]

For the following exercises, solve the equation for x, if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

61. [latex]{\mathrm{log}}_{9}\left(x\right)-5=-4[/latex]

62. [latex]{\mathrm{log}}_{3}\left(x\right)+3=2[/latex]

63. [latex]\mathrm{ln}\left(3x\right)=2[/latex]

64. [latex]\mathrm{ln}\left(x - 5\right)=1[/latex]

65. [latex]\mathrm{log}\left(4\right)+\mathrm{log}\left(-5x\right)=2[/latex]

66. [latex]-7+{\mathrm{log}}_{3}\left(4-x\right)=-6[/latex]

67. [latex]\mathrm{ln}\left(4x - 10\right)-6=-5[/latex]

68. [latex]\mathrm{log}\left(4 - 2x\right)=\mathrm{log}\left(-4x\right)[/latex]

69. [latex]{\mathrm{log}}_{11}\left(-2{x}^{2}-7x\right)={\mathrm{log}}_{11}\left(x - 2\right)[/latex]

70. [latex]\mathrm{ln}\left(2x+9\right)=\mathrm{ln}\left(-5x\right)[/latex]

71. [latex]{\mathrm{log}}_{9}\left(3-x\right)={\mathrm{log}}_{9}\left(4x - 8\right)[/latex]

72. [latex]\mathrm{log}\left({x}^{2}+13\right)=\mathrm{log}\left(7x+3\right)[/latex]

73. [latex]\frac{3}{{\mathrm{log}}_{2}\left(10\right)}-\mathrm{log}\left(x - 9\right)=\mathrm{log}\left(44\right)[/latex]

74. [latex]\mathrm{ln}\left(x\right)-\mathrm{ln}\left(x+3\right)=\mathrm{ln}\left(6\right)[/latex]

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

75. An account with an initial deposit of $6,500 earns 7.25% annual interest, compounded continuously. How much will the account be worth after 20 years?

76. The formula for measuring sound intensity in decibels D is defined by the equation [latex]D=10\mathrm{log}\left(\frac{I}{{I}_{0}}\right)[/latex], where I is the intensity of the sound in watts per square meter and [latex]{I}_{0}={10}^{-12}[/latex] is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of [latex]8.3\cdot {10}^{2}[/latex] watts per square meter?

77. The population of a small town is modeled by the equation [latex]P=1650{e}^{0.5t}[/latex] where t is measured in years. In approximately how many years will the town’s population reach 20,000?

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate x to 3 decimal places.

78. [latex]1000{\left(1.03\right)}^{t}=5000[/latex] using the common log.

79. [latex]{e}^{5x}=17[/latex] using the natural log

80. [latex]3{\left(1.04\right)}^{3t}=8[/latex] using the common log

81. [latex]{3}^{4x - 5}=38[/latex] using the common log

82. [latex]50{e}^{-0.12t}=10[/latex] using the natural log

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

83. [latex]7{e}^{3x - 5}+7.9=47[/latex]

84. [latex]\mathrm{ln}\left(3\right)+\mathrm{ln}\left(4.4x+6.8\right)=2[/latex]

85. [latex]\mathrm{log}\left(-0.7x - 9\right)=1+5\mathrm{log}\left(5\right)[/latex]

86. Atmospheric pressure P in pounds per square inch is represented by the formula [latex]P=14.7{e}^{-0.21x}[/latex], where x is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 pounds per square inch? (Hint: there are 5280 feet in a mile)

87. The magnitude M of an earthquake is represented by the equation [latex]M=\frac{2}{3}\mathrm{log}\left(\frac{E}{{E}_{0}}\right)[/latex] where E is the amount of energy released by the earthquake in joules and [latex]{E}_{0}={10}^{4.4}[/latex] is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing [latex]1.4\cdot {10}^{13}[/latex] joules of energy?

88. Use the definition of a logarithm along with the one-to-one property of logarithms to prove that [latex]{b}^{{\mathrm{log}}_{b}x}=x[/latex].

89. Recall the formula for continually compounding interest, [latex]y=A{e}^{kt}[/latex]. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t such that t is equal to a single logarithm.

90. Recall the compound interest formula [latex]A=a{\left(1+\frac{r}{k}\right)}^{kt}[/latex]. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t.

91. Newton’s Law of Cooling states that the temperature T of an object at any time t can be described by the equation [latex]T={T}_{s}+\left({T}_{0}-{T}_{s}\right){e}^{-kt}[/latex], where [latex]{T}_{s}[/latex] is the temperature of the surrounding environment, [latex]{T}_{0}[/latex] is the initial temperature of the object, and k is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t such that t is equal to a single logarithm.