Đáp án:

Chọn A.

Giải thích các bước giải: \[\begin{array}{*{20}{l}} {{{\log }_4}\left( {{{\log }_2}x} \right) + {{\log }_2}\left( {{{\log }_4}x} \right) = 2}\\ {DK:{\kern 1pt} {\kern 1pt} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}} {x > 0}\\ {{{\log }_2}x > 0}\\ {{{\log }_4}x > 0} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {x > 0}\\ {x > 1} \end{array}} \right. \Leftrightarrow x > 1.}\\ {pt \Leftrightarrow {{\log }_{{2^2}}}\left( {{{\log }_2}x} \right) + {{\log }_2}\left( {{{\log }_{{2^2}}}x} \right) = 2}\\ { \Leftrightarrow \frac{1}{2}{{\log }_2}\left( {{{\log }_2}x} \right) + {{\log }_2}\left( {\frac{1}{2}{{\log }_2}x} \right) = 2}\\ { \Leftrightarrow \frac{1}{2}{{\log }_2}\left( {{{\log }_2}x} \right) + {{\log }_2}\frac{1}{2} + {{\log }_2}\left( {{{\log }_2}x} \right) = 2}\\ { \Leftrightarrow \frac{3}{2}{{\log }_2}\left( {{{\log }_2}x} \right) - 1 = 2}\\ \begin{array}{l} \Leftrightarrow \frac{3}{2}{\log _2}\left( {{{\log }_2}x} \right) = 3\\ \Leftrightarrow {\log _2}\left( {{{\log }_2}x} \right) = 2\\ \Leftrightarrow {\log _2}x = {2^2} = 4 \end{array}\\ { \Leftrightarrow x = {2^4} = 16{\kern 1pt} {\kern 1pt} {\kern 1pt} \left( {tm} \right)}\\ {Vay{\kern 1pt} {\kern 1pt} {\kern 1pt} pt{\kern 1pt} {\kern 1pt} {\kern 1pt} co{\kern 1pt} {\kern 1pt} nghiem{\kern 1pt} {\kern 1pt} {\kern 1pt} x = 16.} \end{array}\]