## Task B-3.1.

Determine all integers $x$ for which

$$
\log _{4}(x+12) \cdot \log _{x} 2 \geqslant 1
$$
## Solution.

The solution of the given inequality $x$ must satisfy the following conditions:

$$
x+12>0 \quad \text{and} \quad x>0, x \neq 1
$$

Since $x \in \mathbb{Z}$, it follows from the above conditions that $x$ is a natural number greater than 1. 1 point The given inequality can be written in the following form

$$
\begin{gathered}
\frac{\log (x+12)}{\log 4} \cdot \frac{\log 2}{\log x} \geqslant 1 \\
\frac{\log (x+12)}{\log x} \geqslant 2
\end{gathered}
$$

Since $\log x>0$ for $x>1$, the given inequality can be multiplied by $\log x$. 1 point Then we have

$$
\begin{aligned}
& \log (x+12) \geqslant \log \left(x^{2}\right) \\
& x+12 \geqslant x^{2}, \\
& x^{2}-x-12 \leqslant 0, \quad \text { i.e. } \quad x \in[-3,4]
\end{aligned}
$$

Given the condition $x>1, x \in \mathbb{Z}$, the solution is $x \in\{2,3,4\}$.