University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 49: 45


(a) $e^{\ln7.2}=7.2$ (b) $e^{-\ln x^2}=\frac{1}{x^2}$ (c) $e^{\ln x-\ln y}=\frac{x}{y}$

Work Step by Step

This whole exercise is based on this property: $$e^{\ln x}=x\hspace{1cm}x\gt0$$ (a) $$e^{\ln7.2}$$ Here, $7.2\gt0$, so $$e^{\ln7.2}=7.2$$ (b) $$e^{-\ln x^2}$$ - Apply Reciprocal Rule for $-\ln x^2$: $-\ln x^2=\ln\frac{1}{x^2}$ $$e^{-\ln x^2}=e^{\ln\frac{1}{x^2}}=\frac{1}{x^2}$$ (c) $$e^{\ln x-\ln y}$$ - Apply Quotient Rule here: $\ln x-\ln y =\ln\frac{x}{y}$ $$e^{\ln x-\ln y}=e^{\ln\frac{x}{y}}=\frac{x}{y}$$
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