Ta có: \(\displaystyle {\log _4}\left\{ {2{{\log }_3}\left[ {1 + {{\log }_2}\left( {1 + 3{{\log }_2}x} \right)} \right]} \right\} = \frac{1}{2}\)
\(\displaystyle \Leftrightarrow 2{\log _3}\left[ {1 + {{\log }_2}\left( {1 + 3{{\log }_2}x} \right)} \right] = {4^{\frac{1}{2}}} = 2\) \(\displaystyle \Leftrightarrow {\log _3}\left[ {1 + {{\log }_2}\left( {1 + 3{{\log }_2}x} \right)} \right] = 1\)
\(\displaystyle \Leftrightarrow 1 + {\log _2}\left( {1 + 3{{\log }_2}x} \right) = 3\) \(\displaystyle \Leftrightarrow {\log _2}\left( {1 + 3{{\log }_2}x} \right) = 2\) \(\displaystyle \Leftrightarrow 1 + 3{\log _2}x = 4 \Leftrightarrow {\log _2}x = 1\)
\(\displaystyle \Leftrightarrow x = 2\).
Chọn B.
