Đáp án: $\,\left[ \begin{array}{l}
x \ge 2\\
0 < x \le 1
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
Đkxđ:{3^x} - 1 > 0 \Rightarrow x > 0\\
{\log _4}\left( {{3^x} - 1} \right).{\log _{\frac{1}{4}}}\left( {\frac{{{3^x} - 1}}{{16}}} \right) \le \frac{3}{4}\\
\Rightarrow {\log _4}\left( {{3^x} - 1} \right).\left[ { - {{\log }_4}\frac{{\left( {{3^x} - 1} \right)}}{{16}}} \right] \le \frac{3}{4}\\
\Rightarrow - {\log _4}\left( {{3^x} - 1} \right).\left( {{{\log }_4}\left( {{3^x} - 1} \right) - {{\log }_4}16} \right) \le \frac{3}{4}\\
\Rightarrow - {\log _4}\left( {{3^x} - 1} \right).\left[ {{{\log }_4}\left( {{3^x} - 1} \right) - 2} \right] \le \frac{3}{4}\\
Dat:{\log _4}\left( {{3^x} - 1} \right) = t\\
\Rightarrow - t\left( {t - 2} \right) \le \frac{3}{4}\\
\Rightarrow {t^2} - 2t + \frac{3}{4} \ge 0\\
\Rightarrow \left[ \begin{array}{l}
t \ge \frac{3}{2}\\
t \le \frac{1}{2}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
{\log _4}\left( {{3^x} - 1} \right) \ge \frac{3}{2}\\
{\log _4}\left( {{3^x} - 1} \right) \le \frac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{3^x} - 1 \ge {4^{\frac{3}{2}}}\\
{3^x} - 1 \le {4^{\frac{1}{2}}}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
{3^x} - 1 \ge 8\\
{3^x} - 1 \le 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le 1
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x \ge 2\\
0 < x \le 1
\end{array} \right.
\end{array}$
