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

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1 Section 1.6 Logarithmic and Exponential Equations Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following: Solve Quadratic Equations (Section 7., pp ) Solve Equations Quadratic in Form (Section 7.4, pp ) Now Work the Are You Prepared? problems on page 816. OBJECTIVES 1 Solve Logarithmic Equations (p. 811) Solve Exponential Equations (p. 813) 3 Solve Logarithmic and Exponential Equations Using a Graphing Utility (p. 815) 1 Solve Logarithmic Equations In Section 1.4 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 11 - x = 3, we use the equivalent exponential equation 1 - x = 3 and solve for x. log 11 - x = x = 3 -x = 7 x = - 7 Change to an exponential expression. Simplify. Divide both sides by. 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 E x a mpl e 1 Solving a L ogarithmic Equation Solve: log 5 x = log 5 9

2 81 CHAPTER 1 Exponential and Logarithmic Functions 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: log 5 x = log 5 9 log 5 x = log 5 9 log a M r = r log a M x = 9 If log a M = log a N, then M = N. x = 3 or x = -3 Recall that the domain of the variable is x 7 0. Therefore, -3 is extraneous and we discard it. Check: log 5 3 log 5 9 The solution set is {3}. log 5 3 log 5 9 log 5 9 = log 5 9 r log a M = log a M r To solve the equation using a graphing utility, graph Y 1 = log 5 x = log x and Y log 5 = log 5 9 = log 9 log 5, and determine the point of intersection. See Figure 47. The point of intersection is (3, ); so x = 3 is the only solution. The solution set is {3}. Figure Now Work p r o b l e m 1 3 E x a mpl e Solving a L ogarithmic Equation Solve: log 5 1x log 5 1x + = 1 The domain of the variable requires that x and x + 7 0, so x 7-6 and x 7 -. This means any solution must satisfy x 7 -. To obtain an exact solution, express the left side as a single logarithm. Then change the equation to exponential form. log 5 1x log 5 1x + = 1 log 5 3 1x + 6 1x + 4 = 1 1x + 6 1x + = 5 1 = 5 x + 8x + 1 = 5 x + 8x + 7 = 0 1x + 7 1x + 1 = 0 x = -7 or x = -1 log a M + log a N = log a 1MN Change to an exponential equation. Simplify. Place the quadratic equation in standard form. Factor. Zero-Product Property Only x = -1 satisfies the restriction that x 7 -, so x = -7 is extraneous. The solution set is {-1}, which you should check. log1x + 6 Graph Y 1 = log 5 1x log 5 1x + = + log 5 log1x + and Y log 5 = 1 and determine the point(s) of intersection. See Figure 48. The point of intersection is 1-1, 1, so x = -1 is the only solution. The solution set is {-1}. Figure 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 Now Work p r o b l e m 1

3 Section 1.6 Logarithmic and Exponential Equations 813 E x a mpl e 3 Solving a L ogarithmic Equation Solve: ln x + ln1x - 4 = ln1x + 6 The domain of the variable requires that x 7 0, x , and x 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 - 4 = ln1x + 6 ln [x1x - 4] = ln1x + 6 In M + ln N = ln1mn x1x - 4 = x + 6 If ln M = ln N, then M = N. x - 4x = x + 6 Simplify. x - 5x - 6 = 0 Place the quadratic equation in standard form. 1x - 6 1x + 1 = 0 Factor. x = 6 or x = -1 Zero-Product Property Since the domain of the variable is x 7 4, discard -1 as extraneous. The solution set is {6}, which you should check. Graph Y 1 = ln x + ln 1x - 4 and Y = ln 1x + 6 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}. Figure 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: Solve: log 3 x = 4 Solution: The domain of the variable x is all real numbers except 0. (a) log 3 x = 4 (b) log 3 x = 4 log a x r = r log a x x = 3 4 = 81 Change to exponential form. log 3 x = 4 Domain of variable is x > 0. x = -9 or x = 9 log 3 x = x = 9 Both 9 and 9 are solutions of log 3 x = 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 1.3 and 1.4, we solved exponential equations algebraically by expressing each side of the equation using the same base. That is, we used the oneto-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 x+1 = 16, notice that 16 = 4 and apply the property above to obtain x + 1 =, from which we find x = 1. 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. E x a mpl e 4 Solving an Exponential Equation Solve: x = 5

4 814 CHAPTER 1 Exponential and Logarithmic Functions Since 5 cannot be written as an integer power of, write the exponential equation as the equivalent logarithmic equation. x = 5 x = log 5 = ln 5 ln c Change@of@Base Formula 110), Section 6.5 Alternatively, we can solve the equation x = 5 by taking the natural logarithm (or common logarithm) of each side. Taking the natural logarithm, x = 5 ln x = ln 5 x ln = ln 5 x = ln 5 ln.3 The solution set is e ln 5 ln f. If M = N, then ln M = ln N. In M r = r ln M Exact solution Approximate solution Graph Y 1 = x and Y = 5 and determine the x- coordinate of the point of intersection. See Figure 50. Figure The approximate solution, rounded to three decimal places, is.3. 4 Now Work p r o b l e m 3 5 E x a mpl e 5 Solving an Exponential Equation Solve: 8 # 3 x = 5 Isolate the exponential expression and then rewrite the statement as an equivalent logarithm. 8 # 3 x = 5 3 x = x = log 3 a 5 lna 8 b = 8 b ln Solve for 3 x. Exact solution Approximate solution Graph Y 1 = 8 # 3 x and Y = 5 and determine the x-coordinate of the point of intersection. See Figure 51. Figure The solution set is e log 3 a 5 8 b f. The approximate solution, rounded to three decimal places, is E x a mpl e 6 Solving an Exponential Equation Solve: 5 x- = 3 3x+ Because the bases are different, we first apply property (7), Section 1.5 (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. Graph Y 1 = 5 x- and Y = 3 3x+ and determine the x-coordinate of the point of intersection. See Figure 5 on the next page.

5 Section 1.6 Logarithmic and Exponential Equations x- = 3 3x+ Figure ln 5 x- = ln 3 3x+ 1x - ln 5 = 13x + ln 3 1ln 5x - ln 5 = 13 ln 3x + ln 3 1ln 5x - 13 ln 3x = ln 3 + ln 5 1ln 5-3 ln 3x = 1ln 3 + ln 5 1ln 3 + ln 5 x = ln 5-3 ln 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 Note that the y-coordinate,.763e 4, is in scientific notation and means.763 * 10-4 = Now Work p r o b l e m 4 1 The next example deals with an exponential equation that is quadratic in form. E x a mpl e 7 Solving an Exponential Equation T hat Is Quadratic in F orm Solve: 4 x - x - 1 = 0 We note that 4 x = 1 x = x = 1 x, so the equation is actually quadratic in form, and we can rewrite it as 1 x - x - 1 = 0 Let u = x ; then u - u - 1 = 0. Graph Y 1 = 4 x - x - 1 and determine the x-intercept. See Figure 53. The x-intercept is, so the solution set is {}. Now we can factor as usual. Figure x x + 3 = 0 x - 4 = 0 or x + 3 = 0 1u - 4 1u + 3 = 0 u - 4 = 0 or u + 3 = 0 x = 4 x = -3 u = x = 4 or u = x = -3 The equation on the left has the solution x =, since x = 4 = ; the equation on the right has no solution, since x 7 0 for all x. The only solution is. The solution set is {} Now Work p r o b l e m 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. E x a mpl e 8 Solution Solving Equations Using a Graphing Utility Solve: x + e x = Express the solution(s) rounded to two decimal places. The solution is found by graphing Y 1 = x + e x and Y =. Since Y 1 is an increasing function (do you know why?), there is only one point of intersection for Y 1 and Y.

6 816 CHAPTER 1 Exponential and Logarithmic Functions Figure 54 4 Y 1 x e x Y Figure 54 shows the graphs of Y 1 and Y. Using the INTERSECT command, the solution is 0.44 rounded to two decimal places. Now Work p r o b l e m 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. 1. Solve x - 7x - 30 = 0. (pp ). Solve 1x x = 0. (pp ) Skill Building 3. Approximate the solution(s) to x 3 = x - 5 using a graphing utility. 4. Approximate the solution(s) to x 3 - x + = 0 using a graphing utility. In Problems 5 3, 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 = 6. log 1x + 6 = 1 7. log 15x = 4 8. log 3 13x - 1 = 9. log 4 1x + = log log 5 1x + 3 = log log 3 x = log log 4 x = log log x = -log log 5 x = 3 log log 1x log 4 = log 3 1x log 3 9 = 17. log x + log1x + 15 = 18. log x + log 1x - 1 = 19. log(x + 1) = 1 + log(x - ) 0. log1x - log1x - 3 = 1 1. log 1x log 1x + 8 = 1. log 6 (x + 4) + log 6 (x + 3) = 1 3. log 8 (x + 6) = 1 - log 8 (x + 4) 4. log 5 1x + 3 = 1 - log 5 1x ln x + ln1x + = 4 6. ln1x ln x = 7. log 3 1x log 3 1x + 4 = 8. log 1x log 1x + 7 = 3 9. log 1>3 1x + x - log 1>3 1x - x = log 4 1x log 4 1x + 3 = log a 1x log a 1x + 6 = log a 1x - - log a 1x log a x + log a 1x - = log a 1x + 4 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. x-5 = x = x = x = x = x = x = x = x = 4 x 4. x+1 = 5 1-x 43. a 3 x 5 b = 7 1-x 44. a 4 1-x 3 b = 5 x x = x x = 1.7 x p 1-x = e x 48. e x+3 = p x 49. x + x - 1 = x + 3 x - = x + 3 x+1-4 = 0 5. x + x+ - 1 = x + 4 x+1-3 = x - 3 x = x - 8 # 5 x = x - 6 # 6 x = # 4 x + 4 # x + 8 = # 49 x + 11 # 7 x + 5 = x - 10 # 4 -x = 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 log 4 1x - = 1 6. log 1x log 6 1x + = 63. e x = -x 64. e x = x e x = x 66. e x = x ln x = -x 68. ln1x = -x ln x = x ln x = -x 71. e x + ln x = 4 7. e x - ln x = e -x = ln x 74. e -x = -ln x

7 Mixed Practice Section 1.6 Logarithmic and Exponential Equations 817 In Problems 75 86, solve each equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places. 75. log 1x log 4 x = log 13x + - log 4 x = log 16 x + log 4 x + log x = 7 [Hint: Change log 4 x to base.] 78. log 9 x + 3 log 3 x = x = x 80. log x log x = ex + e -x = 1 8. ex + e -x = ex - e -x = [Hint: Multiply each side by e x.] 84. ex - e -x = log 5 x + log 3 x = log x + log 6 x = 3 [Hint: Use the Change-of-Base Formula.] 87. f1x = log 1x + 3 and g1x = log 13x + 1. (a) Solve f1x = 3. What point is on the graph of f? (b) Solve g1x = 4. What point is on the graph of g? (c) Solve f1x = g1x. Do the graphs of f and g intersect? If so, where? (d) Solve 1f + g 1x = 7. (e) Solve 1f - g 1x =. 88. f1x = log 3 1x + 5 and g1x = log 3 1x - 1. (a) Solve f1x =. What point is on the graph of f? (b) Solve g1x = 3. What point is on the graph of g? (c) Solve f1x = g1x. Do the graphs of f and g intersect? If so, where? (d) Solve 1f + g 1x = 3. (e) Solve 1f - g 1x =. 89. (a) Graph f1x = 3 x+1 and g(x) = x+, on the same Cartesian plane. (b) Find the point(s) of intersection of the graphs of f and g by solving f1x = g1x. Round answers to three decimal places. Label any intersection points on the graph drawn in part (a). (c) Based on the graph, solve f1x 7 g1x. 90. (a) Graph f1x = 5 x-1 and g(x) = x+1, on the same Cartesian plane. (b) Find the point(s) of intersection of the graphs of f and g by solving f1x = g1x. Label any intersection points on the graph drawn in part (a). (c) Based on the graph, solve f1x 7 g1x. 91. (a) Graph f1x = 3 x and g(x) = 10 on the same Cartesian plane. (b) Shade the region bounded by the y-axis, f1x = 3 x, and g(x) = 10 on the graph drawn in part (a). (c) Solve f1x = g1x and label the point of intersection on the graph drawn in part (a). 9. (a) Graph f1x = x and g(x) = 1 on the same Cartesian plane. (b) Shade the region bounded by the y-axis, f1x = x, and g(x) = 1 on the graph drawn in part (a). (c) Solve f1x = g1x and label the point of intersection on the graph drawn in part (a). 93. (a) Graph f1x = x+1 and g(x) = -x+ on the same Cartesian plane. (b) Shade the region bounded by the y-axis, f1x = x+1, and g(x) = -x+ on the graph draw in part (a). (c) Solve f1x = g1x and label the point of intersection on the graph drawn in part (a). 94. (a) Graph f1x = 3 -x+1 and g1x = 3 x- on the same Cartesian plane. (b) Shade the region bounded by the y-axis, f1x = 3 -x+1, and g1x = 3 x- on the graph draw in part (a). (c) Solve f1x = g1x and label the point of intersection on the graph drawn in part (a). 95. (a) Graph f1x = x - 4. (b) Find the zero of f. (c) Based on the graph, solve f1x (a) Graph g1x = 3 x - 9. (b) Find the zero of g. (c) Based on the graph, solve g1x 7 0. Applications and Extensions 97. A Population Model The resident population of the United States in 010 was 309 million people and was growing at a rate of 0.9% per year. Assuming that this growth rate continues, the model P1t) = ) t 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

8 818 CHAPTER 1 Exponential and Logarithmic Functions 98. A Population Model The population of the world in 011 was 6.91 billion people and was growing at a rate of 1.14% per year. Assuming that this growth rate continues, the model P1t = t-011 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 1 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 V1t = 16, 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 $000? 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 - 1 = Solution A Solution B log 3 1x - 1) = log 3 1x - 1 = 1x - 1 = 3 = 9 log 3 1x - 1 = 1x - 1 = { 3 log 3 1x - 1 = 1 x - 1 = -3 or x - 1 = 3 x - 1 = 3 1 = 3 x = - or x = 4 x = 4 Both solutions given in Solution A check. Explain what caused the solution x = - to be lost in Solution B. Are You Prepared? Answers 1. {-3, 10}. {-, 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 V1t = 16, 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