Task B-3.5. Solve the inequality

$$
x^{\frac{1}{\log x}} \cdot \log x < 1
$$
## Solution.

The inequality makes sense for $x>0$ and $x \neq 1$.

(1 point)

By taking the logarithm of the given inequality, we get

$$
\begin{aligned}
& \frac{1}{\log x} \log x + \log (\log x) < 1 \\
& 1 + \log (\log x) < 1 \\
& \log (\log x) < 0 \\
& \log x < 1 \\
& x < 10^1 \\
& x < 10
\end{aligned}
$$

Considering $x>0$ and $x \neq 1$, the solution to the given inequality is

$$
x \in (0,1) \cup (1, \sqrt[10]{10}) = (0, \sqrt[10]{10}) \setminus \{1\}
$$