5.6 Logarithmic and Exponential Equations

Transcription

1 SECTION 5.6 Logarithmic and Exponential Equations Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following: Solving Equations Using a Graphing Utility (Appendix B, Section B.4, pp. B6 B7) Now Work the Are You Prepared? problems on page 309. OBJECTIVES 1 Solve Logarithmic Equations (p. 305) Solve Exponential Equations (p. 307) Solving Quadratic Equations (Appendix A, Section A.6, pp. A47 A51) 3 Solve Logarithmic and Exponential Equations Using a Graphing Utility (p. 308) 1 Solve Logarithmic Equations In Section 5.4 we solved logarithmic equations by changing a logarithmic expression to an exponential expression. That is, we used the definition of a logarithm: y = log a x is equivalent to x = a y a 7 0, a Z 1 For example, to solve the equation log 11 - x = 3, we write the logarithmic equation as an equivalent exponential equation 1 - x = 3 and solve for x. log 11 - x = x = 3 - x = 7 x =- 7 Change to an exponential statement. Simplify. Solve. 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. 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 Z 1. EXAMPLE 1 Solving a Logarithmic Equation Solve: log 5 x = log 5 9 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 r log a M = log a M r 5 x = log 5 9 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.

2 306 CHAPTER 5 Exponential and Logarithmic Functions Check: log 5 3 log 5 9 log r log a M = log a M r 5 3 log 5 9 log 5 9 = log 5 9 The solution set is 536. Now Work PROBLEM 13 Often we need to use one or more properties of logarithms to rewrite the equation as a single logarithm. In the next example we employ the log of a product property to solve a logarithmic equation. EXAMPLE 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. Solving a Logarithmic 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, we need to express the left side as a single logarithm. Then we will change the equation to an equivalent exponential equation. log 5 1x log 5 1x + = 1 log 5 31x + 61x + 4 = 1 1x + 61x + = 5 1 = 5 x + 8x + 1 = 5 x + 8x + 7 = 0 1x + 71x + 1 = 0 x = - 7 or x = - 1 Change to an exponential statement. 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 5-16, which you should check. Now Work PROBLEM 1 log a M + log a N = log a (MN) EXAMPLE 3 Solving a Logarithmic Equation Solve: ln x = ln1x ln1x - 4 The domain of the variable requires that x 7 0, x , and x As a result, the domain of the variable here is x 7 4. We begin the solution using the log of a difference property. ln x = ln1x ln1x - 4 x + 6 ln x = lna x - 4 b In M - ln N = lna M N b x = x + 6 x - 4 If ln M = ln N, then M = N. x1x - 4 = x + 6 Multiply both sides by x 4. x - 4x = x + 6 Simplify. x - 5x - 6 = 0 Place the quadratic equation in standard form. 1x - 61x + 1 = 0 Factor. x = 6 or x = - 1 Zero-Product Property Since the domain of the variable is x 7 4, we discard - 1 as extraneous. The solution set is {6}, which you should check.

3 SECTION 5.6 Logarithmic and Exponential Equations 307 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 : 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 7 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 solution in part (b) does not find the solution 9 because the domain of the variable was further restricted due to the application of the property log a x r = r log a x. Now Work PROBLEM 31 Solve Exponential Equations In Sections 5.3 and 5.4, 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 Z 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. EXAMPLE 4 Solving Exponential Equations Solve: (a) x = 5 (b) 8 # 3 x = 5 (a) Since 5 cannot be written as an integer power of ( = 4 and 3 = 8), write the exponential equation as the equivalent logarithmic equation. Alternatively, we can solve the equation x = 5 by taking the natural logarithm (or common logarithm) of each side. Taking the natural logarithm, The solution set is e ln 5. ln f x = 5 x = log 5 = ln 5 ln c Change-of-Base Formula (10), Section 5.5 x = 5 ln x = ln 5 x ln = ln 5 x = ln 5 ln L.3 If M = N, then ln M = ln N. In M r = r ln M Exact solution Approximate solution (b) 8 # 3 x = 5 3 x = 5 8 Solve for 3 x.

4 308 CHAPTER 5 Exponential and Logarithmic Functions 5 x = log 3 a 5 lna 8 b = 8 b ln 3 L Exact solution Approximate solution lna 5 8 b The solution set is L ln 3 M. Now Work PROBLEM 35 EXAMPLE 5 EXAMPLE 6 Solving an Exponential Equation Solve: 5 x - = 3 3x + Because the bases are different, we first apply property (7), Section 5.5 (take the natural logarithm of each side), and then use a property of logarithms. The result is an equation in x that we can solve. 5 x - = 3 3x + 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 3 L The solution set is 1ln 3 + ln 5 e ln 5-3 ln 3 f. Now Work PROBLEM 45 If M = N, ln M = ln N. ln M r = r ln M Distribute. Place terms involving x on the left. Factor. Exact solution Approximate solution Solving an Exponential Equation That Is Quadratic in Form Solve: 4 x - x - 1 = 0 We note that 4 x = 1 x = (x) = 1 x, so the equation is quadratic in form, and we can rewrite it as 1 x - x - 1 = 0 Now we can factor as usual. 1 x - 41 x + 3 = 0 x - 4 = 0 or x + 3 = 0 x = 4 x = - 3 Let u = x ; then u - u - 1 = 0. (u - 4)(u + 3) = 0 u - 4 = 0 or u + 3 = 0 u = x = 4 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 56. Now Work PROBLEM 53 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. s for other types are usually studied in calculus, using numerical methods. For such types, we can use a graphing utility to approximate the solution.

5 SECTION 5.6 Logarithmic and Exponential Equations 309 Figure 40 4 EXAMPLE 7 Y 1 x e x Y 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 + 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. Figure 40 shows the graphs of Y 1 and Y. Using the INTERSECT command, the solution is 0.44 rounded to two decimal places. e x Now Work PROBLEM Assess 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. A47 A51) 3. Approximate the solution(s) to x 3 = x - 5 using a graphing. Solve 1x x = 0. (pp. A47 A51) utility. (pp. B6 B7) 4. Approximate the solution(s) to x 3 - x + = 0 using a graphing utility. (pp. B6 B7) Skill Building In Problems 5 3, solve each logarithmic equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places. 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 4 x = log log x = - log 7 log 3 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. log1x + 1 = 1 + log1x - 0. log1x - log1x - 3 = 1 1. log 1x log 1x + 8 = 1. log 6 1x log 6 1x + 3 = 1 3. log 8 1x + 6 = 1 - log 8 1x 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 30. log 4 1x 1>3 1x + x - log 1>3 1x - x = 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. 33. x - 5 = x = x = x = x = x = x = x = a x a 3 x x = 4 x x + 1 = x 3 b = 5 x 5 b = x x = x x = 1.7 x p 1 - x = e x 48. e x + 3 = p x

6 310 CHAPTER 5 Exponential and Logarithmic Functions 49. x + x - 1 = x + 3 x - = x + 3 x = 0 5. x + x = x - 8 # x - 6 # x + 4 x = 0 9 x - 3 x = 0 5 x = x = x - 14 # 3-4 x - 10 # 4-3 # 4 x + 4 # x + 8 = 0 # 49 x + 11 # 7 x + 5 = 0 x = 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 Mixed Practice 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 = 1 [Hint: Change log 4 x to base.] 76. log 13x + - log 4 x = log 16 x + log 4 x + log x = log log x log x 9 x + 3 log 3 x = 14 A 3 B - x = x = e - x e x + = 1 [Hint: Multiply each side by e x.] e x + 8. = e - x e x - e - x = e x = log 5 x + log 3 x = 1 [Hint: Use the Change-of-Base Formula.] e - x 86. log x + log 6 x = 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 + (e) Solve 1f - g1x = 7. g1x =. 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 + (e) Solve 1f - g1x = 3. g1x =. 89. (a) If f1x = 3 x + 1 and g1x = x +, graph f and g on the same (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) If f1x = 5 x - 1 and g1x = x + 1, graph f and g on the same (b) Find the point(s) of intersection of the graphs of f and g by solving f1x = g1x. Label any intersection points (c) Based on the graph, solve f1x 7 g1x. 91. (a) Graph f1x = 3 x and g1x = 10 on the same (b) Shade the region bounded by the y-axis, f1x = 3 x, and g1x = 10 (c) Solve f1x = g1x and label the point of intersection 9. (a) Graph f1x = x and g1x = 1 on the same (b) Shade the region bounded by the y-axis, f1x = x, and g1x = 1 (c) Solve f1x = g1x and label the point of intersection 93. (a) Graph f1x = x + 1 and g1x = - x + on the same (b) Shade the region bounded by the y-axis, f1x = x + 1, and g1x = - x + on the graph draw in part (a). (c) Solve f1x = g1x and label the point of intersection 94. (a) Graph f1x = 3 - x + 1 and g1x = 3 x - on the same (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 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.