3. Solve the inequality $11^{\frac{3 x-4}{4-x}} \cdot 3^{4-x} \leq 99$.
Answer: $x \in\left[4-4 \log _{3} 11 ; 2\right] \cup(4 ;+\infty)$.

Solution. By taking the logarithm of both sides of the inequality with base 3, we get:

$$
\frac{3 x-4}{4-x} \log _{3} 11+(4-x) \leq 2+\log _{3} 11 \Leftrightarrow \frac{4 x-8}{4-x} \log _{3} 11+(2-x) \leq 0 \Leftrightarrow \frac{(x-2)\left(4 \log _{3} 11-4+x\right)}{4-x} \leq 0
$$

Solving the last inequality using the interval method, we find that $x \in\left[4-4 \log _{3} 11 ; 2\right] \cup(4 ;+\infty)$.