338 CHAPTER 5 Exponential and Logarithmic Functions Solving a Logarithmic Equation Solve: x x x ln ln 4 ln 6 ( ) ( ) + − = + Graphing Solution Graph ( ) ( ) = + − y x x x 1 ln ln 4 and ( ) ( ) = + y x x 2 ln 6 using Geogebra, and determine the point(s) of intersection. See Figure 48. The x -coordinate of the point of intersection is 6, so the solution set is { }6 . Now Work PROBLEM 35 EXAMPLE 3 Algebraic Solution The domain of the variable requires that > x 0, − > x 4 0, and + > x 6 0. As a result, the domain of the variable is > x 4. Begin the solution using the log of a product property. ( ) ( ) ( ) [ ] ( ) ( ) ( )( ) + − = + − = + − = + − = + − − = − + = = = − x x x x x x x x x x x x x x x x x x ln ln 4 ln 6 ln 4 ln 6 4 6 4 6 5 6 0 6 1 0 6 or 1 2 2 Because the domain of the variable is x 4, > discard −1 as extraneous. The solution set is { }6 , which you should check. In ( ) + = M N MN ln ln If =M N ln ln , then M N. = Multiply out. Place the quadratic equation in standard form. Factor. Use the Zero-Product Property. Figure 48 2 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: = = > ≠ a a u v a a If , then 0, 1 u v For example, to solve the exponential equation = + 4 16, x2 1 notice that = 16 42 and use the one-to-one property to obtain the equation + = x2 1 2, from which we find = x 1 2 . Not all exponential equations can be readily expressed so that each side of the equation has the same base. For such equations, algebraic techniques often can be CAUTION In using properties of logarithms to solve logarithmic equations, avoid using the property x r x log log , a r a = when r is even. The reason can be seen in this example: Solve: = x log 4 3 2 Solution: The domain of the variable x is all real numbers except 0. (b) = = = = x x x x log 4 2log 4 log 2 9 3 2 3 3 (a) = = = =− = x x x x log 4 3 81 9 or 9 3 2 2 4 Both −9 and 9 are solutions of = x log 4 3 2 (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 0 due to the application of the property = x r x log log . a r a j Change to exponential form. log log x r x a r a = Domain of variable is 0. x >
RkJQdWJsaXNoZXIy NjM5ODQ=