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: 51


$\displaystyle \{\frac{\log 7}{\log 3-\log 5+\log 7} \}$ or $\{1.356\}$

Work Step by Step

... Apply $\log$(...) to both sides and isolate x $ \begin{aligned} \left(\displaystyle \frac{3}{5}\right)^{x}&=7^{1-x}\\ \log\left(\displaystyle \frac{3}{5}\right)^{x}&=\log\left(7^{1-x}\right)\\ x\log(\displaystyle \frac{3}{5})&=(1-x)\log7\\ x\log(\displaystyle \frac{3}{5})&=\log7-x\log7\\ x\log(\displaystyle \frac{3}{5})+x\log7&=\log7\\ x(\log3-\log5+\log7)&=\log7 \end{aligned}$ $x=\displaystyle \frac{\log 7}{\log 3-\log 5+\log 7}\approx 1.356$ Solution set: $\displaystyle \{\frac{\log 7}{\log 3-\log 5+\log 7} \}$ or $\{1.356\}$
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