Answer
The solution set of the equation $\text{ }{{\log }_{2}}x+{{\log }_{2}}\left( x-4 \right)={{\log }_{2}}\left( 3x-10 \right)$ is $\left\{ 5 \right\}$.
Work Step by Step
Consider the expression, $\text{ }{{\log }_{2}}x+{{\log }_{2}}\left( x-4 \right)={{\log }_{2}}\left( 3x-10 \right)$
Apply the product property of the logarithm on the left side of the expression
$\text{ }{{\log }_{2}}x+{{\log }_{2}}\left( x-4 \right)={{\log }_{2}}\left( 3x-10 \right)$
Therefore,
$\begin{align}
& \text{ }{{\log }_{2}}\left( x\left( x-4 \right) \right)={{\log }_{2}}\left( 3x-10 \right) \\
& \text{ }{{\log }_{2}}\left( {{x}^{2}}-4x \right)={{\log }_{2}}\left( 3x-10 \right) \\
& {{x}^{2}}-4x=3x-10 \\
& {{x}^{2}}-7x+10=0
\end{align}$
Solve the above equation,
$\begin{align}
& {{x}^{2}}-2x-5x+10=0 \\
& x\left( x-2 \right)-5\left( x-2 \right)=0 \\
& \left( x-2 \right)\left( x-5 \right)=0
\end{align}$
Further solve the above expression.
$\left( x-2 \right)=0\text{ or }\left( x-5 \right)=0$
When $\left( x-2 \right)=0$, then $ x=2$
When $\left( x-5 \right)=0$, then $ x=5$
If $ x=2$, then substitute this value in equation $\text{ }{{\log }_{2}}x+{{\log }_{2}}\left( x-4 \right)={{\log }_{2}}\left( 3x-10 \right)$,
$\begin{align}
& \text{ }{{\log }_{2}}\left( 2 \right)+{{\log }_{2}}\left( 2-4 \right)={{\log }_{2}}\left( 3\left( 2 \right)-10 \right) \\
& {{\log }_{2}}\left( 2 \right)+{{\log }_{2}}\left( -2 \right)={{\log }_{2}}\left( -4 \right)
\end{align}$
Since the value inside the logarithm cannot be negative, thus $ x=2$ does not satisfy this equation.
If $ x=5$, then substitute this value in equation $\text{ }{{\log }_{2}}x+{{\log }_{2}}\left( x-4 \right)={{\log }_{2}}\left( 3x-10 \right)$,
$\begin{align}
& \text{ }{{\log }_{2}}\left( 5 \right)+{{\log }_{2}}\left( 5-4 \right)={{\log }_{2}}\left( 3\left( 5 \right)-10 \right) \\
& {{\log }_{2}}\left( 5 \right)+{{\log }_{2}}\left( 1 \right)={{\log }_{2}}\left( 5 \right) \\
& {{\log }_{2}}\left( 5 \right)={{\log }_{2}}\left( 5 \right)
\end{align}$
Hence, the solution set of the equation $\text{ }{{\log }_{2}}x+{{\log }_{2}}\left( x-4 \right)={{\log }_{2}}\left( 3x-10 \right)$ is $\left\{ 5 \right\}$.
