## Precalculus (6th Edition) Blitzer

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