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: 49


$\displaystyle \{\frac{\log 3}{\log 36}\}$ or $\{0.307\}$

Work Step by Step

... Apply $\log$(...) to both sides $\log 3^{1-2x}=\log 4^{x}$ $\quad $...Apply$: \quad \log_{a}M^{r}=r\log_{a}M\quad $ $(1-2x)\log 3=x\log 4 \quad $... isolate x $\log 3-2x\log 3=x\log 4$ $\log 3=x\log 4+2x\log 3$ $\log 3=x(\log 4+2\log 3) \quad $...Apply$: \quad \log_{a}M^{r}=r\log_{a}M\quad $ $\log 3=x(\log 4+\log 9) \quad $...Apply$: \quad \log_{a}(MN)=\log_{a}M+\log_{a}N$ $\log 3=x(\log 36)$ $x=\displaystyle \frac{\log 3}{\log 36}\approx 0.307$ Solution set: $\displaystyle \{\frac{\log 3}{\log 36}\}$ or $\{0.307\}$
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