$8.34 \log _{x} \log _{9}\left(3^{x}-9\right)<1$.

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$8.34 \log _{x} \log _{9}\left(3^{x}-9\right)<1$.
8.34 This inequality can be considered under the conditions

$$
\left\{\begin{array} { l } 
{ 3 ^ { x } - 9 > 0 , } \\
{ \operatorname { log } _ { 9 } ( 3 ^ { x } - 9 ) > 0 , }
\end{array} \text { or } \left\{\begin{array} { l } 
{ 3 ^ { x } > 3 ^ { 2 } , } \\
{ 3 ^ { x } - 9 > 1 , }
\end{array} \text { or } \left\{\begin{array}{l}
x>2 \\
3^{x}>10
\end{array}\right.\right.\right.
$$

By taking the logarithm of the second inequality with base 3, we find $x>\log _{3} 10>2$, i.e., in this inequality, the base of the logarithm $x>1$. $^{\prime}$ Therefore, $\log _{9}\left(3^{x}-9\right)0$, which holds for any $y$. Thus, the inequality (1) holds for all $x$, and the solution to the original inequality is the interval $x>\log _{3} 10$.

Answer: $\left(\frac{1}{\lg 3}, \infty\right)$.