Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 6 - Logarithmic Functions - 6.5 Solving Exponential Equations - 6.5 Exercises - Page 528: 66


$h^{-1}(x) = \frac{\ln (\frac{x}{-3.5})}{\ln 1.8}$

Work Step by Step

$h(x) = -3.5(1.8)^{x}$ Let $h(x) = y$ $y = -3.5(1.8)^{x}$ Swap the variables $x$ and $y$ and then solve for $y$ to find the inverse: $x = -3.5(1.8)^{y}$ $(1.8)^{y} = -\frac{x}{3.5}$ $y = \ln_{1.8} (\frac{x}{-3.5})$ $y = \frac{\ln (\frac{x}{-3.5})}{\ln 1.8}$ $h^{-1}(x) = \frac{\ln (\frac{x}{-3.5})}{\ln 1.8}$
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