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.

Using 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 [latex]b[/latex], [latex]S[/latex], and [latex]T[/latex], where [latex]b>0[/latex], [latex]b≠1[/latex], [latex]b^S=b^T[/latex] if and only if [latex]S=T[/latex].

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 [latex]x[/latex], we use the division property of exponents to rewrite the right side so that both sides have the common base, [latex]3[/latex]. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for [latex]x[/latex]:

Rewriting Equations So All Powers Have the Same Base

Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.

For example, consider the equation [latex]256=4^{x−5}[/latex]. We can rewrite both sides of this equation as a power of [latex]2[/latex]. Then we apply the rules of exponents, along with the one-to-one property, to solve for [latex]x[/latex]:

Solving Exponential Equations Using Logarithms

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since [latex]log(a)=log(b)[/latex] is equivalent to [latex]a=b[/latex], we may apply logarithms with the same base on both sides of an exponential equation.

Equations Containing e

One common type of exponential equations are those with base [latex]e[/latex]. This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base [latex]e[/latex] on either side, we can use the natural logarithm to solve it.

Extraneous Solutions

Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation [latex]log_b(x)=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]log_2(2)+log_2(3x−5)=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 [latex]x[/latex]:

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers [latex]x>0[/latex], [latex]S>0[/latex], [latex]T>0[/latex] and any positive real number [latex]b[/latex], where [latex]b≠1[/latex],

For example,

So, if [latex]x−1=8[/latex], then we can solve for [latex]x[/latex], and we get [latex]x=9[/latex]. To check, we can substitute [latex]x=9[/latex] into the original equation: [latex]log_2(9−1)=log_2(8)=3[/latex]. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation [latex]log(3x−2)−log(2)=log(x+4)[/latex]. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for [latex]x[/latex]:

To check the result, substitute [latex]x=10[/latex] into [latex]log(3x−2)−log(2)=log(x+4)[/latex].

Solving Applied Problems Using Exponential and Logarithmic Equations

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. Table 1 lists the half-life for several of the more common radioactive substances.

Table 1

Substance	Use	Half-life
gallium-67	nuclear medicine	80 hours
cobalt-60	manufacturing	5.3 years
technetium-99m	nuclear medicine	6 hours
americium-241	construction	432 years
carbon-14	archeological dating	5,715 years
uranium-235	atomic power	703,800,000 years

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

where

• [latex]A_0[/latex] is the amount initially present
• [latex]T[/latex] is the half-life of the substance
• [latex]t[/latex] is the time period over which the substance is studied
• [latex]A(t)[/latex] is the amount of the substance present after time [latex]t[/latex]

Media

Access these online resources for additional instruction and practice with exponential and logarithmic equations.

• Solving Logarithmic Equations
• Solving Exponential Equations with Logarithms

Section Exercises

Verbal

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?

Algebraic

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

4. [latex]4^{−3v−2}=4^{−v}[/latex]

5. [latex]64⋅4^{3x}=16[/latex]

6. [latex]3^{2x+1}⋅3x=243[/latex]

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

8. [latex]625⋅5^{3x+3}=125[/latex]

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

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

For the following exercises, use logarithms to solve.

11. [latex]9^{x−10}=1[/latex]

12. [latex]2e^{6x}=13[/latex]

13. [latex]e^{r+10}−10=−42[/latex]

14. [latex]2⋅10^{9a}=29[/latex]

15. [latex]−8⋅10^{p+7}−7=−24[/latex]

16. [latex]7e^{3n−5}+5=−89[/latex]

17. [latex]e^{−3k}+6=44[/latex]

18. [latex]−5e^{9x−8}−8=−62[/latex]

19. [latex]−6e^{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]7e^{8x+8}−5=−95[/latex]

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

24. [latex]4e^{3x+3}−7=53[/latex]

25. [latex]8e^{−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]3e^{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]log(\frac{1}{100})=−2[/latex]

30. [latex]log_{324}(18)=\frac{1}{2}[/latex]

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

31. [latex]5log_7n=10[/latex]

32. [latex]−8log_9x=16[/latex]

33. [latex]4+log_2(9k)=2[/latex]

34. [latex]2log(8n+4)+6=10[/latex]

35. [latex]10−4ln(9−8x)=6[/latex]

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

36. [latex]ln(10−3x)=ln(−4x)[/latex]

37. [latex]log_{13}(5n−2)=log_{13}(8−5n)[/latex]

38. [latex]log(x+3)−log(x)=log(74)[/latex]

39. [latex]ln(−3x)=ln(x^2−6x)[/latex]

40. [latex]log_4(6−m)=log_43m[/latex]

41. [latex]ln(x−2)−ln(x)=ln(54)[/latex]

42. [latex]log_9(2n^2−14n)=log_9(−45+n^2)[/latex]

43. [latex]ln(x^2−10)+ln(9)=ln(10)[/latex]

For the following exercises, solve each equation for [latex]x[/latex].

44. [latex]log(x+12)=log(x)+log(12)[/latex]

45. [latex]ln(x)+ln(x−3)=ln(7x)[/latex]

46. [latex]log_2(7x+6)=3[/latex]

47. [latex]ln(7)+ln(2−4x^2)=ln(14)[/latex]

48. [latex]log_8(x+6)−log_8(x)=log_8(58)[/latex]

49. [latex]ln(3)−ln(3−3x)=ln(4)[/latex]

50. [latex]log_3(3x)−log_3(6)=log_3(77)[/latex]

Graphical

For the following exercises, solve the equation for [latex]x[/latex], 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.

51. [latex]log_9(x)−5=−4[/latex]

52. [latex]log_3(x)+3=2[/latex]

53. [latex]ln(3x)=2[/latex]

54. [latex]ln(x−5)=1[/latex]

55. [latex]log(4)+log(−5x)=2[/latex]

56. [latex]−7+log_3(4−x)=−6[/latex]

57. [latex]ln(4x−10)−6=−5[/latex]

58. [latex]log(4−2x)=log(−4x)[/latex]

59. [latex]log_{11}(−2x^2−7x)=log_{11}(x−2)[/latex]

60. [latex]ln(2x+9)=ln(−5x)[/latex]

61. [latex]log_9(3−x)=log_9(4x−8)[/latex]

62. [latex]log(x^2+13)=log(7x+3)[/latex]

63. [latex]\frac{3}{log_2(10)}−log(x−9)=log(44)[/latex]

64. [latex]ln(x)−ln(x+3)=ln(6)[/latex]

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

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

66. The formula for measuring sound intensity in decibels [latex]D[/latex] is defined by the equation [latex]D=10log(\frac{I}{I_0})[/latex], where [latex]I[/latex] 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⋅10^2[/latex] watts per square meter?

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

Extensions

68. Use the definition of a logarithm along with the one-to-one property of logarithms to prove that [latex]b^{log_bx}=x[/latex].

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

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

71. Newton’s Law of Cooling states that the temperature [latex]T[/latex] of an object at any time t can be described by the equation [latex]T=T_s+(T_0−T_s)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 [latex]k[/latex] is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time [latex]t[/latex] such that [latex]t[/latex] is equal to a single logarithm.