9.084. $\log _{x}\left(\log _{9}\left(3^{x}-9\right)\right)<1$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \in(0 ; 1) \cup(1 ;+\infty), \\ 3^{x}-9>1,\end{array}\left\{\begin{array}{l}x \in(0 ; 1) \cup(1 ;+\infty) \\ 3^{x}>10,\end{array} x \in\left(\log _{3} 10 ;+\infty\right)\right.\right.$.

We obtain the following system of inequalities

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x > \log _ { 3 } 1 0 , } \\
{ \log _ { x } ( \log _ { 9 } ( 3 ^ { x } - 9 ) ) \log _{3} 10, \\
\log _{9}\left(3^{x}-9\right) \log _ { 3 } 1 0 , } \\
{ \log _ { 9 } ( 3 ^ { x } - 9 )  \log _ { 3 } 1 0 , } \\
{ 3 ^ { x } - 9 \log _{3} 10, \\
\left(3^{x}\right)^{2}-3^{x}+9>0
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x \in\left(\log _{3} 10 ;+\infty\right), \\
x \in \mathbb{R} .
\end{array}\right.
\end{aligned}
$$

From this, $x \in\left(\frac{1}{\lg 3} ;+\infty\right)$.

Answer: $\quad x \in\left(\frac{1}{\lg 3} ;+\infty\right)$