Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Section 3.3 - Properties of Logarithms - Exercise Set - Page 475: 26

Answer

$2-\displaystyle \frac{1}{2}\log_{8}(x+1)$

Work Step by Step

$\displaystyle \log_{8}(\frac{64}{\sqrt{x+1}})=$ ...apply the Quotient Rule: $\displaystyle \quad \log_{b}(\frac{M}{N})=\log_{b}\mathrm{M}-\log_{b}\mathrm{N}$ $\log_{8}64-\log_{8}\sqrt{x+1}$ ... write $64$ as $8^{2},\quad\sqrt{x+1}=(x+1)^{1/2}$ $=\log_{8}8^{2}-\log_{8}(x+1)^{1/2}$ $\quad $...apply the Power Rule: $\quad \log_{b}(M^{p})=p\cdot\log_{b}\mathrm{M}$ $=2\displaystyle \log_{8}8-\frac{1}{2}\log_{8}(x+1)$ $\quad $...apply the basic property: $\log_{b}b=1$ $=2-\displaystyle \frac{1}{2}\log_{8}(x+1)$
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