ĐK: \(\displaystyle \left\{ \begin{array}{l}x > 0\\{\log _4}2x \ne 0\\{\log _{16}}8x \ne 0\end{array} \right.\).
Khi đó, phương trình \(\displaystyle \frac{{{{\log }_2}x}}{{{{\log }_4}2x}} = \frac{{{{\log }_8}4x}}{{{{\log }_{16}}8x}}\)\(\displaystyle \Leftrightarrow {\log _2}x.{\log _{16}}8x = {\log _4}2x.{\log _8}4x\)
\(\displaystyle \Leftrightarrow {\log _2}x.\frac{1}{4}{\log _2}8x\) \(\displaystyle = \frac{1}{2}{\log _2}2x.\frac{1}{3}{\log _2}4x\)
\(\displaystyle \Leftrightarrow 3{\log _2}x.\left( {3 + {{\log }_2}x} \right)\)\(\displaystyle = 2\left( {1 + {{\log }_2}x} \right)\left( {2 + {{\log }_2}x} \right)\)
\(\displaystyle \Leftrightarrow 9{\log _2}x + 3\log _2^2x\) \(\displaystyle = 2\left( {2 + 3{{\log }_2}x + \log _2^2x} \right)\)
\(\displaystyle \Leftrightarrow \log _2^2x + 3{\log _2}x - 4 = 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}{\log _2}x = 1\\{\log _2}x = - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2\\x = \frac{1}{{16}}\end{array} \right.\left( {TM} \right)\)
Vậy tập nghiệm \(\displaystyle \left\{ {2;\frac{1}{{16}}} \right\}\).
Chọn D.
