College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.6 - Logarithmic and Exponential Equations - 6.6 Assess Your Understanding - Page 465: 55


$\displaystyle \{\frac{\ln\pi}{1+\ln\pi}\}$ or $\{0.534\}$

Work Step by Step

... Apply $\ln$(...) to both sides and isolate x $\ln\pi^{1-x}=\ln e^{x}$ $(1-x)\ln\pi=x$ $\ln\pi-x\ln\pi=x$ $\ln\pi=x+x\ln\pi$ $\ln\pi=x(1+\ln\pi)$ $x=\displaystyle \frac{\ln\pi}{1+\ln\pi}\approx 0.534$ The solution set is $\displaystyle \{\frac{\ln\pi}{1+\ln\pi}\}$ or $\{0.534\}$
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