452 CHAPTER 4 Inverse, Exponential, and Logarithmic Functions Exponential Equations Because the graph of ƒ1x2 = ax is that of a oneto-one function, to solve ax1 = ax2, we need only show that x 1 = x2. This property is used to solve an exponential equation, which is an equation with a variable as exponent. EXAMPLE 3 Graphing Reflections and Translations Graph each function. Show the graph of y = 2x for comparison. Give the domain and range. (a) ƒ1x2 = -2x (b) ƒ1x2 = 2x+3 (c) ƒ1x2 = 2x-2 - 1 SOLUTION In each graph, we show in particular how the point 10, 12 on the graph of y = 2x has been translated. (a) The graph of ƒ1x2 = -2x is that of ƒ1x2 = 2x reflected across the x-axis. See Figure 18. The domain is 1-∞, ∞2, and the range is 1-∞, 02. (b) The graph of ƒ1x2 = 2x+3 is the graph of ƒ1x2 = 2x translated to the left 3 units, as shown in Figure 19. The domain is 1-∞, ∞2, and the range is 10, ∞2. (c) The graph of ƒ1x2 = 2x-2 - 1 is that of ƒ1x2 = 2x translated to the right 2 units and down 1 unit. See Figure 20. The domain is 1-∞, ∞2, and the range is 1-1, ∞2. EXAMPLE 4 Solving an Exponential Equation Solve A1 3B x = 81. SOLUTION Write each side of the equation using a common base. a 1 3b x = 81 13-12x = 81 Definition of negative exponent 3-x = 81 1am2n = amn 3-x = 34 Write 81 as a power of 3. -x = 4 Set exponents equal (Property (b)). x = -4 Multiply by -1. CHECK a 1 3b x = a 1 3b -4 = 34 = 81 ✓ The solution set is 5-46. S Now Try Exercise 73. −100 −5 100 1 The x-intercept of the graph of y = A1 3B x - 81 can be used to verify the solution in Example 4. x y 2 f(x) = –2x (0, 21) (0, 1) –2 –2 2 –4 0 y = 2x Figure 18 (23, 1) (0, 1) y x –6 –3 2 2 4 6 8 f(x) = 2x+3 0 y = 2x Figure 19 y 4 y 5 21 0 4 x y = 2x (2, 0) (0, 1) f(x) = 2x22 2 1 Figure 20 S Now Try Exercises 39, 43, and 47.
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