Answer
The required value is $ x=5$
Work Step by Step
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.
