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Câu hỏi: Cho số thực x thỏa mãn ${{\log _2}\left( {{{\log }_4}x} \right) = {\log _4}\left( {{{\log }_2}x} \right) + m}$. Tính giá trị của ${{\log _2}x}$ theo ${m}$.
A. ${{2^{m + 1}}}$.
B. ${{4^{m + 1}}}$.
C. ${{m^2}}$.
D. ${{4^m}}$.
A. ${{2^{m + 1}}}$.
B. ${{4^{m + 1}}}$.
C. ${{m^2}}$.
D. ${{4^m}}$.
Ta có ${{\log }_{2}}\left( {{\log }_{4}}x \right)={{\log }_{4}}\left( {{\log }_{2}}x \right)+m$
$\Leftrightarrow {{\log }_{2}}\left( \dfrac{1}{2}{{\log }_{2}}x \right)=\dfrac{1}{2}{{\log }_{2}}\left( {{\log }_{2}}x \right)+m$
$\Leftrightarrow -1+{{\log }_{2}}\left( {{\log }_{2}}x \right)=\dfrac{1}{2}{{\log }_{2}}\left( {{\log }_{2}}x \right)+m$
$\Leftrightarrow \dfrac{1}{2}{{\log }_{2}}\left( {{\log }_{2}}x \right)=m+1$
$\Leftrightarrow {{\log }_{2}}\left( {{\log }_{2}}x \right)=2\left( m+1 \right)$
$\Leftrightarrow {{\log }_{2}}x={{2}^{2\left( m+1 \right)}}={{4}^{m+1}}$
$\Leftrightarrow {{\log }_{2}}\left( \dfrac{1}{2}{{\log }_{2}}x \right)=\dfrac{1}{2}{{\log }_{2}}\left( {{\log }_{2}}x \right)+m$
$\Leftrightarrow -1+{{\log }_{2}}\left( {{\log }_{2}}x \right)=\dfrac{1}{2}{{\log }_{2}}\left( {{\log }_{2}}x \right)+m$
$\Leftrightarrow \dfrac{1}{2}{{\log }_{2}}\left( {{\log }_{2}}x \right)=m+1$
$\Leftrightarrow {{\log }_{2}}\left( {{\log }_{2}}x \right)=2\left( m+1 \right)$
$\Leftrightarrow {{\log }_{2}}x={{2}^{2\left( m+1 \right)}}={{4}^{m+1}}$
Đáp án B.