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: 52


$\displaystyle \{\frac{\log 4-\log 3}{\log 5+\log 4-\log 3} \}$ or $\{0.152\}$

Work Step by Step

... Apply $\log$(...) to both sides and isolate x $\displaystyle \log\left(\frac{4}{3}\right)^{1-x}=\log\left(5^{x}\right)$ $(1-x)\displaystyle \log(\frac{4}{3})=x\log 5$ $\displaystyle \log(\frac{4}{3})-x\log(\frac{4}{3})=x\log 5$ $\displaystyle \log(\frac{4}{3})=x\log 5+x\log(\frac{4}{3})$ $\log 4-\log 3=x(\log 5+\log 4-\log 3)$ $x=\displaystyle \frac{\log 4-\log 3}{\log 5+\log 4-\log 3}\approx 0.152$ Solution set: $\displaystyle \{\frac{\log 4-\log 3}{\log 5+\log 4-\log 3} \}$ or $\{0.152\}$
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