Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.6 Logarithmic and Exponential Equations - 4.6 Assess Your Understanding - Page 337: 55

Answer

$x =\dfrac{\ln{\pi}}{1+\ln{\pi}}\approx 0.534$

Work Step by Step

To find the value of $x$, we take the $\log$ of both sides and then isolate $x$ as follows: $$\ln (\pi^{1-x})=\ln (e^{x})$$ Recall : $\log m^n= n \log m$ Use the rule above to obtain: $(1-x)(\ln\pi)=x \\ \ln{\pi}-x\cdot \ln{\pi}=x\\ \ln\pi=x+x\cdot\ln\pi$ Factor out $x$ on the right side, then solve for $x$ to obtain: $\ln\pi=x(1+\ln\pi)\\ \dfrac{\ln\pi}{1+\ln\pi}=x\\ x \approx 0.534$ Thus, our answer is: $x =\dfrac{\ln{\pi}}{1+\ln{\pi}}\approx 0.534$
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