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: 35

Answer

Solution set: $ \{-2+4\sqrt{2}\}$ or $\{3.6569\}$

Work Step by Step

(*) For the equation to be defined, it must be true that $\left\{x+2 \gt 0 \right.$ Thus, $x \gt -2 $ ...Apply$: \quad \log_{a}M^{r}=r\log_{a}M\quad $ $\log_{6}(x+2)^{2}=\log_{6}2^{3}+\log_{6}4$ ...Apply$: \quad \log_{a}(MN)=\log_{a}M+\log_{a}N$ $\log_{6}(x+2)^{2}=\log_{6}(2^{3}\cdot 4)$ ...Apply$: \quad $If $\log_{a}M=\log_{a}N,$ then $M=N$ $(x+2)^{2}=32$ $x+2=\pm\sqrt{2^{5}}$ $x=-2\pm 4\sqrt{2}$ The solutions need to satisfy (*), so $ x=-2-4\sqrt{2}\approx-7.6569$ is eliminated. Solution set: $ \{-2+4\sqrt{2}\}$ or $\{3.6569\}$
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