## Precalculus (6th Edition) Blitzer

The solution set is $\left( 67 \right)$
${{\log }_{4}}\left( {{x}^{2}}-9 \right)-{{\log }_{4}}\left( x+3 \right)-{{\log }_{4}}64=0$ So, \begin{align} & {{\log }_{4}}\left( \frac{{{x}^{2}}-9}{x+3} \right)={{\log }_{4}}64 \\ & \frac{{{x}^{2}}-9}{x+3}=64 \\ & {{x}^{2}}-9=64\left( x+3 \right) \\ & {{x}^{2}}-9=64x+192 \end{align} \begin{align} & {{x}^{2}}-64x+201=0 \\ & \left( x-67 \right)\left( x+3 \right)=0 \\ & x-67=0,x+3=0 \\ & x=67,x=-3 \end{align} We reject the solution $x=-3$ because it requires us to take the log of $0$, which is undefined. Hence, the solution set is $\left( 67 \right)$