3. Solve the inequality $2^{\frac{1-x}{3+x}} \cdot 7^{3+x} \geq 56$.
Answer: $x \in(-3 ;-2] \cup\left[4 \log _{7} 2-3 ;+\infty\right)$.

Solution. By taking the logarithm of both sides of the inequality with base 7, we get:

$$
\frac{1-x}{3+x} \log _{7} 2+(3+x) \geq 1+3 \log _{7} 2 \Leftrightarrow \frac{-4 x-8}{3+x} \log _{7} 2+(2+x) \geq 0 \Leftrightarrow \frac{(x+2)\left(-4 \log _{7} 2+3+x\right)}{3+x} \geq 0
$$

Solving the last inequality using the interval method, we find that $x \in(-3 ;-2] \cup\left[4 \log _{7} 2-3 ;+\infty\right)$.