Solve the inequality

$$
\log _{x}\left(2.5-\frac{1}{x}\right)>1
$$
Solution. From the definition of the logarithm, it follows that $x>0, x \neq 1$ and $2.5-\frac{1}{x}>0$ implies $x>\frac{2}{5}$. Let's rewrite the right side of the inequality in logarithmic form:

$$
\log _{x}\left(2.5-\frac{1}{x}\right)>\log _{x} x
$$

We know that the logarithmic function is strictly monotonic, and if the base is greater than 1, it is strictly increasing, and if it is less than 1, it is decreasing. First, let's examine the inequality when $\frac{2}{5}<x<1$. In this case, the inequality $2.5-\frac{1}{x}<x$ leads to the inequality $2 x^{2}-5 x+2>0$. The roots of the quadratic equation $2 x^{2}-5 x+2=0$ are

$$
x_{1}=\frac{5+\sqrt{25-16}}{4}=2, \quad x_{2}=\frac{1}{2}
$$

The function is positive when $x<\frac{1}{2}$ or $x>2$, comparing this with the conditions on $x$, the solution to the inequality is: $\frac{2}{5}<x<\frac{1}{2}$.

If $x>1$, then $2.5-\frac{1}{x}>x$. Rearranging the inequality:

$$
2 x^{2}-5 x+2<0
$$

which holds if $\frac{1}{2}<x<2$. In this case, considering the condition, the solution to the inequality is: $1<x<2$.

Therefore, the inequality holds for the elements of the set $\left(\frac{2}{5} ; \frac{1}{2}\right) \cup(1 ; 2)$.