Task B-3.2. (20 points) Solve the inequality

$$
\log _{x} 2 \cdot \log _{2 x} 2 \cdot \log _{2}(16 x)<1
$$
Solution. For the inequality to make sense, the following must be satisfied:

$$
x>0, \quad x \neq 1 \quad \text { and } \quad x \neq \frac{1}{2}
$$

Now we have:

$$
\begin{aligned}
\frac{1}{\log _{2} x} \cdot \frac{1}{\log _{2}(2 x)} \cdot \log _{2}(16 x) & <1 \\
\frac{1}{\log _{2} x} \cdot \frac{1}{1+\log _{2} x} \cdot\left(4+\log _{2} x\right) & <1
\end{aligned}
$$

With the substitution

$$
\log _{2} x=t
$$

we get

$$
\begin{aligned}
\frac{4+t}{t(1+t)} & <1 \\
\frac{4-t^{2}}{t(1+t)} & <0
\end{aligned}
$$

The solution to this inequality is

$$
t \in\langle-\infty,-2\rangle \cup\langle-1,0\rangle \cup\langle 2, \infty\rangle
$$

(3 points)

From (1), (2), and (3), we get $x \in\left\langle 0, \frac{1}{4}\right\rangle \cup\left\langle\frac{1}{2}, 1\right\rangle \cup\langle 4, \infty\rangle$.

(5 points)