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: 48


(a) $\ln(e^{\sec\theta})=\sec\theta$ (b) $\ln e^{e^x}=e^x$ (c) $\ln (e^{2\ln x})=2\ln x$

Work Step by Step

This whole exercise is based on this property: $$\ln e^x=x\hspace{1cm}x\gt0$$ (a) $$\ln(e^{\sec\theta})=\ln e^{\sec\theta}=\sec\theta$$ Therefore, $$\ln(e^{\sec\theta})=\sec\theta$$ (b) $$\ln e^{e^x}=\ln e^{(e^x)}=e^x$$ Therefore, $$\ln e^{e^x}=e^x$$ (c) $$\ln (e^{2\ln x})=\ln e^{2\ln x}=2\ln x$$ Therefore, $$\ln (e^{2\ln x})=2\ln x$$
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