Precalculus (6th Edition) Blitzer

The solution set of the equation $\text{ }{{\log }_{2}}x+{{\log }_{2}}\left( x-4 \right)={{\log }_{2}}\left( 3x-10 \right)$ is $\left\{ 5 \right\}$.
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 $\left\{ 5 \right\}$.