9.060. $2 \log _{8}(x-2)-\log _{8}(x-3)>\frac{2}{3}$.
## Solution.

## Domain of the function: $\quad x>3$.

## We have

$$
\log _{8}(x-2)^{2}-\log _{8}(x-3)>\frac{2}{3}, \log _{8} \frac{(x-2)^{2}}{x-3}>\frac{2}{3} \Leftrightarrow\left\{\begin{array}{l}
\frac{(x-2)^{2}}{x-3}>8^{\frac{2}{3}} \\
x>3
\end{array} \Leftrightarrow\right.
$$

$$
\Leftrightarrow\left\{\begin{array} { l } 
{ \frac { x ^ { 2 } - 4 x + 4 } { x - 3 } - 4 > 0 } \\
{ x > 3 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \frac { ( x - 4 ) ^ { 2 } } { x - 3 } > 0 } \\
{ x > 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \neq 4 \\
x>3
\end{array}\right.\right.\right.
$$

Answer: $\quad x \in(3 ; 4) \cup(4 ; \infty)$.