Đáp án:
\[\frac{{3 - x}}{{4\left( {3 + x} \right)}}\]
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
x = {\log _{12}}27 = {\log _{12}}{3^3} = 3.{\log _{12}}3 = \frac{3}{{{{\log }_3}12}}\\
= \frac{3}{{{{\log }_3}{{3.2}^2}}} = \frac{3}{{{{\log }_3}3 + 2{{\log }_3}2}} = \frac{3}{{1 + 2{{\log }_3}2}}\\
\Rightarrow 1 + 2{\log _3}2 = \frac{3}{x}\\
\Rightarrow {\log _3}2 = \frac{{\frac{3}{x} - 1}}{2} = \frac{{3 - x}}{{2x}} \Rightarrow {\log _2}3 = \frac{{2x}}{{3 - x}}
\end{array}\]
Suy ra
\[\begin{array}{l}
{\log _6}16 = {\log _6}{2^4} = 4.{\log _6}2 = \frac{4}{{{{\log }_2}6}} = \frac{4}{{{{\log }_2}2 + {{\log }_2}3}}\\
= \frac{4}{{1 + {{\log }_2}3}} = \frac{4}{{1 + \frac{{2x}}{{3 - x}}}} = \frac{4}{{\frac{{3 + x}}{{3 - x}}}} = \frac{{3 - x}}{{4\left( {3 + x} \right)}}
\end{array}\]
