Answer
The required value is $ x=4$
Work Step by Step
Consider the given equation,
${{\log }_{4}}\left( 2x+1 \right)={{\log }_{4}}\left( x-3 \right)+{{\log }_{4}}\left( x+5 \right)$
Apply the product rule of logarithms
$\begin{align}
& {{\log }_{4}}\left( 2x+1 \right)={{\log }_{4}}\left( x-3 \right)+{{\log }_{4}}\left( x+5 \right) \\
& {{\log }_{4}}\left( 2x+1 \right)={{\log }_{4}}\left( x-3 \right)\left( x+5 \right) \\
\end{align}$
Comparing both sides, we get,
$\begin{align}
& \left( 2x+1 \right)=\left( x-3 \right)\left( x+5 \right) \\
& 2x+1={{x}^{2}}+5x-3x-15 \\
\end{align}$
Now, add to both sides $-2x-1$,
$\begin{align}
& 2x+1-2x-1={{x}^{2}}+5x-3x-15-2x-1 \\
& {{x}^{2}}-16=0 \\
& \left( x+4 \right)\left( x-4 \right)=0
\end{align}$
The roots of the above equation are
$ x=4,-4$
Here $ x=-4$ is not the solution of the given expression because
$\begin{align}
& {{\log }_{4}}\left( 2\cdot \left( -4 \right)+1 \right)={{\log }_{4}}\left( -4-3 \right)+{{\log }_{4}}\left( -4+5 \right) \\
& {{\log }_{4}}\left( -7 \right)={{\log }_{4}}\left( -7 \right)+{{\log }_{4}}\left( 1 \right) \\
\end{align}$
So, by definition of ${{\log }_{a}}b $, $ b>0$ but $-7$ is not greater than zero
Hence $ x=-4$ does not satisfy the equation ${{\log }_{4}}\left( 2x+1 \right)={{\log }_{4}}\left( x-3 \right)+{{\log }_{4}}\left( x+5 \right)$
Thus $ x=4$ is the only solution of the given expression.
