3. Solve the inequality $17^{\frac{5 x-3}{3-x}} \cdot 2^{3-x} \leq 68$.

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Answer: $x \in\left[3-6 \log _{2} 17 ; 1\right] \cup(3 ;+\infty)$.

Solution. By taking the logarithm of both sides of the inequality with base 2, we get:

$$
\frac{5 x-3}{3-x} \log _{2} 17+(3-x) \leq 2+\log _{2} 17 \Leftrightarrow \frac{6 x-6}{3-x} \log _{2} 17+(1-x) \leq 0 \Leftrightarrow \frac{(x-1)\left(6 \log _{2} 17-3+x\right)}{3-x} \leq 0
$$

Solving the last inequality using the interval method, we find that $x \in\left[3-6 \log _{2} 17 ; 1\right] \cup(3 ;+\infty)$.