Math 147 - Assignment 7 Solutions - Fall 2012 - BSU - Jaimos F Skriletz 1. Inverse Functions x+3 Consider the function f (x) = x−2 (a) Find the inverse function f −1 (x). y = f (x) = x + 3x − 2 y(x − 2) = x + 3 yx − 2y = x + 3 yx − x = 2y + 3 x(y − 1) = 2y + 3 x = f −1 (y) = 2y + 3 y−1 2y + 3 2x + 3 or you could say f −1 (x) = y−1 x−1 (b) To check this is an inverse function notice that Thus x = f −1 (y) = 2x + 3 +3 2x + 3 x−1 f (f −1 )(x) = f = 2x + 3 x−1 −2 x−1   2x + 3 + 3  x−1  x−1   =  x−1 2x + 3 −2 x−1 (2x + 3) + 3(x − 1) = (2x + 3) − 2(x − 1) 2x + 3 + 3x − 3 5x = = =x 2x + 3 − 2x + 2 5 And also x+3 2 +3 x+3 x−2 −1 −1 f (f (x)) = f = x+3 x−2 −1 x−2   x+3 2 + 3   x−2 x−2  =  x+3  x−2 −1 x−2 2(x + 3) + 3(x − 2) = (x + 3) − (x − 2) 2x + 6 + 3x − 6 5x = = =x x+3−x+2 5 Thus f (f −1 (x)) = x and f −1 (f (x)) = x. 1 Math 147 - Assignment 7 Solutions - Fall 2012 - BSU - Jaimos F Skriletz x+3 (c) The domain of f (x) = is x − 2 6= 0 or x 6= 2. x−2 2x + 3 is x − 1 6= 0 or x 6= 1. The domain of f −1 (x) = x−1 Since the range of f (x) is the domain of f −1 (x) and visa versa, we get that x+3 • f (x) = has domain x−2 {x | x 6= 2} = (−∞, 2) ∪ (2, ∞) 2 And has the range 2x + 3 • f −1 (x) = has domain x−1 {x | x 6= 1} = (−∞, 1) ∪ (1, ∞) {x | x 6= 1} = (−∞, 1) ∪ (1, ∞) {x | x 6= 2} = (−∞, 2) ∪ (2, ∞) And has the range 2. Logarithms Calculate out the following logarithms by first rewriting the equation in exponential form: (a) y = log6 (216) 6y = 6log6 (216) = 216 6y = 63 y=3 Thus log6 (216) = 3 (b) y = log5 log5 5y = 5 1 625 ! 1 625 = 1 625 1 = 5−4 54 y = −4 5y = Thus log5 (1/625) = −4 (c) y = ln(e25 ) ey = 3ln(e 25 = e25 ey = e25 y = 25 Thus ln(e25 ) = 25. Another method is since ln(x) is the inverse function of ex , we have the inverse function property ln(ex ) = x. Thus ln(e25 ) = 25 Math 147 - Assignment 7 Solutions - Fall 2012 - BSU - Jaimos F Skriletz 3. Logarithmic Functions 3 Consider the function f (x) = ln(3 − x) + 5 = ln(−(x − 3)). (a) The parent function is p(x) = ln(x) and this function is obtained from ln(x) by • Horizontal reflection about the y-axis. • Move 3 units to the right. • Move 5 units up. (b) The domain is all reals such that 3 − x > 0 or x < 3. The range is all real numbers (since the parent function ln(x) has all real numbers as the range and shifting all real numbers up 5 is still all real numbers). The vertical asymptote occurs when 3 − x = 0 or x = 3. (c) The graph of f (x) is 4. Properties of Logarithms (a) Use Properties of logarithms to expand 3x2 ln = ln(3x2 ) − ln((x + 1)10 ) = ln(3) + ln(x2 ) − 10 ln(x + 1) = ln(3) + 2 ln(x) − 10 ln(x + 1) (x + 1)10 (b) Use Properties of logarithms to combine and simplify 3 log(x) − log(x2 − 1) + 2 log(x − 1) = log(x3 ) + log((x − 1)2 ) − log(x2 − 1) 3 x (x − 1)2 = log x2 − 1 3 x (x − 1)2 = log (x − 1)(x + 1) 3 x (x − 1) = log x+1 Math 147 - Assignment 7 Solutions - Fall 2012 - BSU - Jaimos F Skriletz 5. Exponential and Logarithmic Equations 4 Solve the following equations: (a) 21−x = 52x ln(21−x ) = ln(52x ) (1 − x) ln(2) = 2x ln(5) ln(2) − x ln(2) = 2x ln(5) ln(2) = 2x ln(5) + x ln(2) ln(2) = x(2 ln(5) + ln(2)) x= ln(2) ln(2) ln(2) = = ≈ 0.177 2 ln(5) + ln(2) ln(52 · 2) ln(50) (b) 10 =2 1 + e−x 10 = 2(1 + e−x ) 5 = 1 + e−x 4 = e−x ln(4) = −x x = − ln(4) ≈ −1.386 (c) e2x − ex − 12 = 0 (ex )2 − (ex ) − 12 = 0 (ex − 4)(ex + 3) = 0 ex = 4 x = ln(4) or ex = −3 or No Solutions Notice that since the range of ex is (0, ∞), there are no solutions to ex = −3. Thus the only solution is x = ln(4) ≈ 1.386 (d) log2 (x + 2) + log2 (x − 1) = 2 log2 ((x + 2)(x − 1)) = 2 (x + 2)(x − 1) = 22 = 4 x2 + x − 2 = 4 x2 + x − 6 = 0 (x + 3)(x − 2) = 0 x = −3 or x = 2 Since x = −3 is not in the domain of log2 (x + 2) + log2 (x − 1) it is an extraneous solution. Thus the only solution is x = 2.