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SECTION 4.6 4.6 Logarithmic and Exponential Equations 309 Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following: • Solving Equations Using a Graphing Utility (Section 1.3, pp. 24–26) • Solving Quadratic Equations (Appendix, Section A.5, pp. 988–996) Now work the ‘Are You Prepared?’ problems on page 313. OBJECTIVES 1 2 3 Solve Logarithmic Equations Using the Properties of Logarithms Solve Exponential Equations Solve Logarithmic and Exponential Equations Using a Graphing Utility 1 Solve Logarithmic Equations Using the Properties ✓ of Logarithms In Section 4.4 we solved logarithmic equations by changing a logarithm to exponential form. Often, however, some manipulation of the equation (usually using the properties of logarithms) is required before we can change to exponential form. 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. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall 310 Exponential and Logarithmic Functions CHAPTER 4 EXAMPLE 1 Solving a Logarithmic Equation Solve: 2 log 5 x = log5 9 Algebraic Solution Graphing Solution Because each logarithm is to the same base, 5, we can obtain an exact solution as follows: To solve the equation using a graphing 2 log x utility, graph Y1 = 2 log5 x = and log 5 2 log 5 x log 5 x2 x2 x = 3 or ✔ CHECK: = = = x 2 log 5 3 log 5 9 log 5 32 log 5 9 log5 9 = log 5 9 log 5 9 log 5 9 9 = -3 loga Mr = r loga M If loga M = loga N, then M = N. Recall that logarithms of negative numbers are not defined, so, in the expression 2 log5 x, x must be positive. Therefore, -3 is extraneous and we discard it. log 9 , and determine the point log 5 of intersection. See Figure 45. Y2 = log 5 9 = Figure 45 4 4 r loga M = loga M r 6 The solution set is 536. 4 NOW WORK PROBLEM EXAMPLE 2 9. Solving a Logarithmic Equation Solve: log 41x + 32 + log 412 - x2 = 1 Algebraic Solution Graphing Solution To obtain an exact solution, we need to express the left side as a single logarithm. Then we will change the expression to exponential form. Graph Y1 = log 41x + 32 + log 412 - x2 = log41x + 32 + log 412 - x2 = 1 log 431x + 3212 - x24 = 1 1x + 3212 - x2 = 4 = 4 2 -x - x + 6 = 4 x2 + x - 2 = 0 1x + 221x - 12 = 0 x = -2 or x = 1 1 loga M + loga N = loga(MN) Change to an exponential expression. log1x + 32 log12 - x2 and Y2 = 1 and + log 4 log 4 determine the points of intersection. See Figure 46. Figure 46 2 Simplify. Place the quadratic equation in standard form. 3 Factor. Zero-Product Property 3 1 Since the arguments of each logarithmic expression in the equation are positive for both x = -2 and x = 1, neither is extraneous. We leave the check to you. 2 3 The solution set is 5-2, 16. NOW WORK PROBLEM 3 1 13. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall SECTION 4.6 Logarithmic and Exponential Equations 311 2 Solve Exponential Equations ✓ In Sections 4.3 and 4.4, we solved certain exponential equations algebraically by expressing each side of the equation with the same base. However, many exponential equations cannot be rewritten so each side has the same base. In such cases, properties of logarithms along with algebraic techniques can sometimes be used to obtain a solution. EXAMPLE 3 Solving an Exponential Equation Solve: 4 x - 2 x - 12 = 0 Algebraic Solution We note that 4 = 12 2 = 2 = 12 2 , so the equation is actually quadratic in form, and we can rewrite it as 2 x x 2x 12 x2 - 2 x - 12 = 0 2 x 2 Graphing Solution Graph Y1 = 4 x - 2 x - 12 and determine the x-intercept. See Figure 47. Figure 47 Let u = 2x; then u2 - u - 12 = 0. 100 Now we can factor as usual. 12 x - 4212 x + 32 = 0 (u - 4)(u + 3) = 0 2 - 4 = 0 or 2x + 3 = 0 u - 4 = 0 or u + 3 = 0 x x 2 = 4 2 = -3 u = 2x = 4 u = 2x = -3 x The equation on the left has the solution x = 2, since 2 = 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 526. x 1 4 30 In the algebraic solution of the previous example, we were able to write the exponential expression using the same base after utilizing some algebra, obtaining an exact solution to the equation. When this is not possible, logarithms can sometimes be used to obtain the solution. EXAMPLE 4 Solving an Exponential Equation Solve: 2x = 5 Algebraic Solution Graphing Solution Since 5 cannot be written as an integral power of 2, we write the exponential equation as the equivalent logarithmic equation. 2x = 5 ln 5 x = log 2 5 = ln 2 q Graph Y1 = 2 x and Y2 = 5 and determine the x-coordinate of the point of intersection. See Figure 48. Figure 48 16 Change-of-Base Formula (9), Section 4.5 Alternatively, we can solve the equation 2 x = 5 by taking the natural logarithm (or common logarithm) of each side. Taking the natural logarithm, 2x = 5 ln 2 x = ln 5 If M = N, then ln M = ln N. x ln 2 = ln 5 ln Mr = r ln M ln 5 Exact solution x = ln 2 Approximate solution L 2.322 NOW WORK PROBLEM 1 4 4 The approximate solution, rounded to three decimal places, is 2.322. 21. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall 312 Exponential and Logarithmic Functions CHAPTER 4 EXAMPLE 5 Solving an Exponential Equation Solve: 8 # 3x = 5 Algebraic Solution Graphing Solution 8 # 3x = 5 3x = 5 8 Graph Y1 = 8 # 3x and Y2 = 5 and determine the xcoordinate of the point of intersection. See Figure 49. Solve for 3x. 5 lna b 5 8 x = log 3 a b = 8 ln 3 Figure 49 24 Exact solution 2 L -0.428 Approximate solution 1 8 The approximate solution, rounded to three decimal places, is –0.428. EXAMPLE 6 Solving an Exponential Equation Solve: 5x - 2 = 33x + 2 Algebraic Solution Graphing Solution Because the bases are different, we first apply Property (6), Section 4.5 (take the natural logarithm of each side), and then use appropriate properties of logarithms. The result is an equation in x that we can solve. Graph Y1 = 5x - 2 and Y2 = 33x + 2 and determine the x-coordinate of the point of intersection. See Figure 50. 5x - 2 ln 5x - 2 1x - 22 ln 5 1ln 52x - 2 ln 5 1ln 52x - 13 ln 32x 1ln 5 - 3 ln 32x 33x + 2 ln 33x + 2 13x + 22 ln 3 13 ln 32x + 2 ln 3 2 ln 3 + 2 ln 5 21ln 3 + ln 52 21ln 3 + ln 52 x = ln 5 - 3 ln 3 L -3.212 = = = = = = Figure 50 If M = N, ln M = ln N. 0.0015 ln Mr = r ln M Distribute. Place terms involving x on the left. Factor. 4 Exact solution Approximate solution 0 0.0005 The approximate solution, rounded to three decimal places, is –3.212. NOW WORK PROBLEM 29. 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. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall SECTION 4.6 EXAMPLE 7 Figure 51 Logarithmic and Exponential Equations 313 Solving Equations Using a Graphing Utility Solve: x + ex = 2 Express the solution(s) rounded to two decimal places. 4 Y1 x e x The solution is found by graphing Y1 = x + ex and Y2 = 2. Y1 is an increasing function (do you know why?), and so there is only one point of intersection for Y1 and Y2 . Figure 51 shows the graphs of Y1 and Y2 . Using the INTERSECT command, the solution is 0.44 rounded to two decimal places. Solution Y2 2 0 1 0 NOW WORK PROBLEM 53. Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall