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