Tìm tập hợp nghiệm của phương trình \(\frac{{{{\log }_2}x}}{{{{\log }_4}2x}} = \frac{{{{\log }_8}4x}}{{{{\log }_{16}}8x}}\)
A. \({\left\{ 2 \right\}}\)
B. \({\left\{ {\frac{1}{4}} \right\}}\)
C. \({\left\{ {2;\frac{1}{4}} \right\}}\)
D. \({\left\{ {2;\frac{1}{{16}}} \right\}}\)
\(\begin{array}{l}
\frac{{{{\log }_2}x}}{{{{\log }_4}2x}} = \frac{{{{\log }_8}4x}}{{{{\log }_{16}}8x}}\,\,\,\left( {x > 0} \right)\\
\Rightarrow {\log _2}x.{\log _{16}}8x = {\log _8}4x.{\log _4}2x\\
\Leftrightarrow {\log _2}x.\left[ {\frac{1}{4}.(3 + {{\log }_2}x)} \right] = \frac{1}{3}.(2 + {\log _2}x).\left[ {\frac{1}{2}.(1 + {{\log }_2}x)} \right]\\
\Leftrightarrow 3{\log _2}x(3 + {\log _2}x) = 2(2 + {\log _2}x)(1 + {\log _2}x)\\
\Leftrightarrow 3\log _2^2x + 9{\log _2}x = 2\log _2^2x + 6{\log _2}x + 4\\
\Leftrightarrow \log _2^2x + 3{\log _2}x - 4 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\log _2}x = 1\\
{\log _2}x = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \frac{1}{{16}}
\end{array} \right.
\end{array}\)
Chọn B.
-- Mod Toán 12
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