ĐK: $x>0$ ta có: $PT\Leftrightarrow -{{\log }_{4}}({{3}^{x}}-1).{{\log }_{4}}\frac{{{3}^{x}}-1}{16}\le \frac{3}{4}$

$\Leftrightarrow {{\log }_{4}}({{3}^{x}}-1)\left[ {{\log }_{4}}({{3}^{x}}-1)-2 \right]\ge -\frac{3}{4}$ . Đặt$t={{\log }_{4}}({{3}^{x}}-1)$ ta có: ${{t}^{2}}-2t+\frac{3}{4}\ge 0\Leftrightarrow \left[ \begin{align} & t\ge \frac{3}{2} \\ & t\le \frac{1}{2} \\ \end{align} \right.$

$\Rightarrow \left[ \begin{align} & {{\log }_{4}}({{3}^{x}}-1)\ge \frac{3}{2} \\ & {{\log }_{4}}({{3}^{x}}-1)\le \frac{1}{2} \\ \end{align} \right.$.. $\Leftrightarrow \left[ \begin{align} & {{3}^{x}}-1\ge 8 \\ & {{3}^{x}}-1\le 2 \\ \end{align} \right.$$\Leftrightarrow \left[ \begin{align} & x\ge 2 \\ & x\le 1 \\ \end{align} \right.$ Kết hợp ĐK ban đầu x > 0