9.300 Solve: $\log _{x^{2}-3} 729>3$.
## Solution.

$$
\begin{aligned}
& \log _{x^{2}-3} 9^{3}>3 \Leftrightarrow \log _{x^{2}-3} 9>\log _{x^{2}-3}\left(x^{2}-3\right) \Leftrightarrow \\
& \Leftrightarrow\left[\begin{array} { l } 
{ \{ \begin{array} { l } 
{ 0 < x^2 - 3 < 1 , } \\
{ x^2 - 3 > 1 , } \\
{ 0 < 9 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
1<x^{2}-3<9, \\
\varnothing
\end{array} \Leftrightarrow 2<|x|<\sqrt{12} .\right.\right.
\end{aligned}
$$

Answer: $x \in(-\sqrt{12} ;-2) \cup(2 ; \sqrt{12})$.