9.026. $\log _{0.3}(3 x-8)>\log _{0.3}\left(x^{2}+4\right)$.
Solution.

This inequality is equivalent to the system of inequalities

$$
\left\{\begin{array} { l } 
{ 3 x - 8 > 0 } \\
{ 3 x - 8 \frac{8}{3} \\
x^{2}-3 x+12>0
\end{array}\right.\right.
$$

Since $x^{2}-3 x+12>0$ for $x \in R$, the last system of inequalities is equivalent to the inequality $x>\frac{8}{3}$.

Answer: $\quad x \in\left(\frac{8}{3} ; \infty\right)$