ĐKXĐ: $\left\{ \begin{align} & x>0;x\ne 1 \\ & {{9}^{x}}-72>0 \\ & {{\log }_{3}}({{9}^{x}}-72)>0 \\ & {{\log }_{x}}({{\log }_{3}}({{9}^{x}}-72))\ne 0 \\ \end{align} \right.$$<=>\left\{ \begin{align} & x>0;x\ne 1 \\ & {{9}^{x}}-72>0 \\ & {{9}^{x}}-72>1 \\ & {{\log }_{3}}({{9}^{2}}-72)\ne 1 \\ \end{align} \right.$$<=>\left\{ \begin{align} & {{9}^{x}}>73 \\ & {{9}^{x}}-72\ne 3 \\ \end{align} \right.$$<=>\left\{ \begin{align} & x>{{\log }_{9}}73 \\ & x\ne {{\log }_{9}}75 \\ \end{align} \right.(*)$.

Hàm số xác định khi : $\frac{1}{{{\log }_{x}}({{\log }_{3}}({{9}^{x}}-72))}-1\ge 0<=>\frac{1-{{\log }_{x}}(lo{{g}_{3}}({{9}^{x}}-72))}{{{\log }_{x}}({{\log }_{3}}({{9}^{x}}-72))}\ge 0<=>\left\{ \begin{align} & {{\log }_{x}}({{\log }_{3}}({{9}^{x}}-72))>0 \\ & {{\log }_{x}}({{\log }_{3}}({{9}^{x}}-72))\le 1 \\ \end{align} \right.$

Vì x thỏa mãn (*) nên hệ BPT trên $<=>\left\{ \begin{align} & {{\log }_{3}}({{9}^{x}}-72)>1 \\ & {{\log }_{3}}({{9}^{x}}-72)\le x \\ \end{align} \right.$$<=>\left\{ \begin{align} & {{9}^{x}}-72>3 \\ & {{9}^{x}}-72\le {{3}^{x}} \\ \end{align} \right.$ $<=>\left\{ \begin{align} & {{9}^{x}}>75 \\ & ({{3}^{x}}-9)({{3}^{x}}+8)\le 0 \\ \end{align} \right.$

$<=>\left\{ \begin{align} & {{9}^{x}}>75 \\ & {{3}^{x}}\le 9 \\ \end{align} \right.$$<=>\left\{ \begin{align} & x>{{\log }_{9}}75 \\ & x\le 2 \\ \end{align} \right.=>{{\log }_{9}}75