## Precalculus (6th Edition) Blitzer

The required value is $x=5$
Consider the given equation, ${{\log }_{2}}\left( x+3 \right)+{{\log }_{2}}\left( x-3 \right)=4$ Apply the product rule of logarithm \begin{align} & {{\log }_{2}}\left( x+3 \right)+{{\log }_{2}}\left( x-3 \right)=4 \\ & {{\log }_{2}}\left( x+3 \right)\left( x-3 \right)=4 \\ \end{align} The provided equation can be written as \begin{align} & {{\log }_{2}}\left( x+3 \right)\left( x-3 \right)=4{{\log }_{2}}2 \\ & {{\log }_{2}}\left( x+3 \right)\left( x-3 \right)={{\log }_{2}}{{2}^{4}} \\ \end{align} Comparing both sides, \begin{align} & \left( x+3 \right)\left( x-3 \right)={{2}^{4}} \\ & {{x}^{2}}-9=16 \end{align} Now, adding $9$ on both sides, we get, \begin{align} & {{x}^{2}}-9+9=16+9 \\ & {{x}^{2}}=25 \end{align} Now, take the square root of both sides, \begin{align} & \sqrt{{{x}^{2}}}=\sqrt{25} \\ & x=\pm 5 \end{align} Here $x=-5$ is not the solution of the given expression because \begin{align} & {{\log }_{2}}\left( -5+3 \right)+{{\log }_{2}}\left( -5-3 \right)=4 \\ & {{\log }_{2}}\left( -2 \right)+{{\log }_{2}}\left( -8 \right)=4 \end{align} By definition of ${{\log }_{a}}b$, $b>0$ but $-2$ and $-8$ is not greater than zero. Thus, $x=-5$ does not satisfy the equation ${{\log }_{2}}\left( x+3 \right)+{{\log }_{2}}\left( x-3 \right)=4$ Therefore, $x=5$ is the only solution.