Precalculus (6th Edition) Blitzer

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

Chapter 11 - Cumulative Review Exercises - Page 1180: 5


The solution of the equation ${{\log }_{4}}\left( {{x}^{2}}-9 \right)-{{\log }_{4}}\left( x+3 \right)=3$ is $67$.

Work Step by Step

Consider the provided equation, ${{\log }_{4}}\left( {{x}^{2}}-9 \right)-{{\log }_{4}}\left( x+3 \right)=3$ Using, for $ a>0$ and $ b>0$, ${{\log }_{b}}a-{{\log }_{b}}c={{\log }_{b}}\left( \frac{a}{c} \right)$ $\begin{align} & {{\log }_{4}}\left( \frac{{{x}^{2}}-9}{x+3} \right)=3 \\ & {{\log }_{4}}\left( \frac{\left( x-3 \right)\left( x+3 \right)}{x+3} \right)=3 \\ & {{\log }_{4}}\left( x-3 \right)=3 \end{align}$ Using,if for $ a>0$ and $ b>0$, ${{\log }_{b}}a=c $, then ${{b}^{c}}=a $ $\begin{align} & {{\log }_{4}}\left( x-3 \right)=3 \\ & x-3={{4}^{3}} \\ & x-3=64 \\ & x=67 \end{align}$ Thus, the solution of the equation ${{\log }_{4}}\left( {{x}^{2}}-9 \right)-{{\log }_{4}}\left( x+3 \right)=3$ is $ x=67$
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