Bài giải

Điều kiện:$0

PT $\Leftrightarrow {{\log }_{x}}2-{{\log }_{16}}x=0\Leftrightarrow {{\log }_{x}}2-{{\log }_{{{2}^{4}}}}x=0\Leftrightarrow {{\log }_{x}}2-\frac{1}{4}{{\log }_{2}}x=0$

$\Leftrightarrow {{\log }_{x}}2-\frac{1}{4{{\log }_{x}}2}=0\Leftrightarrow \frac{4{{({{\log }_{x}}2)}^{2}}-1}{4{{\log }_{x}}2}=0\Leftrightarrow 4{{({{\log }_{x}}2)}^{2}}-1=0$

$\Leftrightarrow {{({{\log }_{x}}2)}^{2}}=\frac{1}{4}$$\Leftrightarrow \left[ \begin{align}& {{\log }_{x}}2=\frac{1}{2} \\ & {{\log }_{x}}2=-\frac{1}{2} \\ \end{align} \right.$

 $\Leftrightarrow \left[ \begin{align}& 2={{x}^{\frac{1}{2}}} \\ & 2={{x}^{-\frac{1}{2}}} \\ \end{align} \right.$

$\Leftrightarrow \left[ \begin{align}& {{x}_{1}}=4 \\ & {{x}_{2}}=\frac{1}{4} \\ \end{align} \right.$

Vậy ${{x}_{1}}.{{x}_{2}}=4.\frac{1}{4}=1$