340 CHAPTER 5 Exponential and Logarithmic Functions Algebraic Solution Note that ( ) ( ) = = = 4 2 2 2 , x x x x 2 2 2 so the equation is quadratic in form and can be written as ( ) − − = 2 2 12 0 x x 2 Let =u 2 ; x then − − = u u 12 0. 2 Now we can factor as usual. ( )( ) − + = − = + = = = − 2 4 2 3 0 2 4 0 or 2 3 0 2 4 2 3 x x x x x x The equation on the left has the solution = x 2, since = = 2 4 2 ; x 2 the equation on the right has no solution, since > 2 0 x for all x. The only solution is 2. The solution set is { }2 . EXAMPLE 6 Figure 51 The approximate solution, rounded to three decimal places, is −3.212. Solving an Exponential Equation Solve: = − + 5 3 x x 2 3 2 Algebraic Solution There are two exponential expressions where the bases are different, so first take the natural logarithm of both sides, and then use the fact that ln = M r M ln r .This results in an equation we know how to solve. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = = − = + − = + − = + − = + = + − ≈ − − + − + x x x x x x x x 5 3 ln5 ln3 2 ln5 3 2 ln3 ln5 2ln 5 3ln3 2ln3 ln5 3ln3 2ln3 2ln5 ln5 3ln3 2 ln3 ln5 2 ln3 ln5 ln5 3 ln3 3.212 x x x x 2 3 2 2 3 2 The solution set is { } ( ) + − 2 ln3 ln5 ln5 3ln3 . Graphing Solution Graph ( ) = − y x 1 5x 2 and ( ) = + y x 2 3 x3 2 using Geogebra, and determine the x-coordinate of the point of intersection. See Figure 51. Now Work PROBLEM 57 NOTE Because of the properties of logarithms, exact solutions involving logarithms often can be expressed in multiple ways. For example, the solution to 5 3 x x 2 3 2 = − + from Example 6 can be expressed equivalently as − 2 ln 15 ln 5 ln 27 or as ln 225 ln 5 27 , ( ) among others. Do you see why? j The next example deals with an exponential equation that is quadratic in form. Now Work PROBLEM 65 Solving an Exponential Equation That Is Quadratic in Form Solve: − − = 4 2 12 0 x x Graphing Solution Graph = − − Y 4 2 12 x x 1 , and determine the x-intercept. See Figure 52. The x-intercept is 2, so the solution set is { }2 . If = = M N M N , ln ln . = M r M ln ln r Distribute. The equation is now linear in x. Place terms involving x on the left. Factor. Exact solution Approximate solution Figure 52 100 230 21 4 Y1 5 4 x 2 2 x 2 12 EXAMPLE 7 ( )( ) − + = u u 4 3 0 − = + = u u 4 0 or 3 0 = = = =− u u 2 4 or 2 3 x x
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