Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.6 - Exponential and Logarithmic Equations; Further Applications - 9.6 Exercises - Page 630: 25

Answer

$x=-\pi$

Work Step by Step

Using the properties of logarithms, the given equation, $ \ln e^{-x}=\pi $ is equivalent to \begin{align*}\require{cancel} (-x)\ln e&=\pi &(\text{use }\log_b x^y=y\log_b x) \\ (-x)(1)&=\pi &(\text{use }\ln e=1) \\ -x&=\pi \\ x&=-\pi .\end{align*} Hence, the solution to the equation $ \ln e^{-x}=\pi $ is $ x=-\pi $.
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