Bài giải

${{\log }_{2}}(x+3)+{{\log }_{2}}(x-1)={{\log }_{2}}5$

$\Leftrightarrow {{\log }_{2}}\text{ }\!\![\!\!\text{ }(x+3)(x-1)\text{ }\!\!]\!\!\text{ }={{\log }_{2}}5$

$\Leftrightarrow \left\{ \begin{align}& x-1>0 \\ & (x+3)(x-1)=5 \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>1 \\ & {{x}^{2}}+2x-8=0 \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>1 \\& x=-8\vee x=2 \\ \end{align} \right.\Rightarrow x=2$