Given the real number $s$. Solve the inequality

$$
\log _{\frac{1}{s}} \log _{s} x>\log _{s} \log _{s} x
$$
Solution. From the definition of the logarithm, it follows that $s>0, s \neq 1$, as well as $x>0$ and $\log _{s} x>0$. Let's convert the left side of (1) to a logarithm with base $s$:

$$
\log _{\frac{1}{s}} \log _{s} x=\frac{\log _{s} \log _{s} x}{\log _{s} \frac{1}{s}}=-\log _{s} \log _{s} x
$$

Thus, we can write the inequality (1) in the following form:

$$
-\log _{s} \log _{s} x>\log _{s} \log _{s} x
$$

From this, $\log _{s} \log _{s} x1$, the logarithm function is strictly monotonically increasing, $\log _{s} x$ is 0 at 1, and 1 at $s, \log _{s} \log _{s} x1$.

If $s0$. The solutions to the inequality are: for $s>1$, $1<x<s$; for $0<s<1$, $0<x<s$.