Ta có: ${{\log }_{2}}x+3{{\log }_{x}}2=4\Leftrightarrow {{\log }_{2}}x+\frac{3}{{{\log }_{2}}x}=4\left( 1\ne x>0 \right)\xrightarrow{t={{\log }_{2}}x}t+\frac{3}{t}=4$

$\Leftrightarrow {{t}^{2}}-4t+3=0$ $\Leftrightarrow \left[ \begin{align} & t=1 \\ & t=3 \\ \end{align} \right.$ $\Rightarrow \left[ \begin{align} & {{\log }_{2}}x=1 \\ & {{\log }_{2}}x=3 \\ \end{align} \right.$ $\Leftrightarrow \left[ \begin{align} & x=2 \\ & x=8 \\ \end{align} \right.$