304 CHAPTER 5 Exponential and Logarithmic Functions 4 Solve Exponential Equations Equations that involve terms of the form > ≠ a a a , where 0 and 1, x are referred to as exponential equations . Such equations can sometimes be solved by using the Laws of Exponents and property (3): Solution Begin with the graph of = y e .x Figure 31 shows the steps. Figure 31 (0, 21) (1, 2e) (2, 2e2) x y 23 26 3 (3, 21) (4, 2e) (5, 2e2) x y 23 26 (1, e) x y 3 6 3 (2, e2) (0, 1) (c) y 5 2ex23 y 5 0 (b) y 5 2ex (a) y 5 ex y 5 0 Replace x by x 23; shift right 3 units. Multiply by 21; reflect about the x-axis. y 5 0 ( ) –2, 1 –e2 ( ) –2, –1 –e2 ( ) 1, –1 –e2 ( ) –1, 1 –e ( ) –1, – 1 –e ( ) 2, – 1 –e As Figure 31(c) illustrates, the domain of ( ) = − − f x ex 3 is the interval ( ) −∞ ∞, , and the range is the interval ( ) −∞, 0 . The horizontal asymptote is the line = y 0. The y -intercept is ( ) = − = − ≈ − − − f e e 0 0.05. 0 3 3 Check: Graph = − − Y ex 1 3 to verify the graph obtained in Figure 31(c). Now Work PROBLEM 59 = = a a u v If , then . u v (3) Property (3) is a consequence of the fact that exponential functions are one-toone.To use property (3), each side of the equality must be written with the same base. In Words When two exponential expressions with the same base are equal, then their exponents are equal. Solving an Exponential Equation Solve: = − 4 8 x x 2 1 EXAMPLE 7 Figure 32 Algebraic Solution Graphing Solution Graph ( ) = − y x 1 4 x2 1 and ( ) = y x 2 8 .x Use Geogebra to determine the point of intersection. See Figure 32. Write each exponential expression so each has the same base. The graphs intersect at ( ) 2, 64 , so the solution set is { }2 . x x x x 4 8 2 2 2 2 1 3 4 2 3 2 x x x 2 1 2 2 1 2 2 1 3 ( ) 22 ( ) 23 ( ) x2 1 = ( ) 2 1 x 2 1 The solution set is { }2 .} Now Work PROBLEMS 69 AND 79 = = 4 2 ; 8 2 2 3 If = a a , u v then =u v. ( ) = a a r s rs
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