9.024. $\log _{3} \frac{3 x-5}{x+1} \leq 1$.
Solution.

The inequality is equivalent to the system of inequalities

$$
\left\{\begin{array} { l } 
{ \frac { 3 x - 5 } { x + 1 } > 0 , } \\
{ \frac { 3 x - 5 } { x + 1 } \leq 3 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ ( x - \frac { 5 } { 3 } ) ( x + 1 ) > 0 , } \\
{ - \frac { 1 5 } { x + 1 } \leq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left(x-\frac{5}{3}\right)(x+1)>0 \\
x>-1
\end{array}\right.\right.\right.
$$

Using the interval method, we find $x>\frac{5}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-525.jpg?height=134&width=614&top_left_y=989&top_left_x=333)

Answer: $\quad x \in\left(\frac{5}{3} ; \infty\right)$