Example 11. Solve the inequality

$$
\sqrt{\frac{9}{10}} \cdot\left(\frac{9}{10}\right)^{x-1}>\frac{10^{(3 / 4) x-1}}{\sqrt{10}}
$$
Solution. Both sides of the given inequality are positive for any $x$. Taking the logarithm of it to the base 10, we obtain the inequality

$$
(x-1) \lg \frac{9}{10}+\frac{1}{2} \lg \frac{9}{10}>\left(\frac{3}{4} x-1\right)-\frac{3}{2}
$$

i.e., the inequality

$$
x\left(\lg \frac{9}{10}-\frac{3}{4}\right)>\frac{1}{2} \lg \frac{9}{10}-\frac{3}{2}
$$

equivalent to inequality (8). Since $\lg \frac{9}{10}-\frac{3}{4}<0$, then

$$
x<\frac{1}{2} \frac{\lg \frac{9}{10}-3}{\lg \frac{9}{10}-\frac{3}{4}}
$$

i.e.,

$$
x<4 \frac{\lg 3-2}{8 \lg 3-7}
$$

Thus, the interval $\left(-\infty ; \frac{4(\lg 3-2)}{8 \lg 3-7}\right)$ is the set of all solutions to inequality (8).

Solving exponential inequalities that have three different bases, which are consecutive terms of a geometric progression, and where these bases are raised to the same power, which depends on $x$, can be reduced by a change of variable to solving a quadratic inequality.

Such exponential inequalities, in particular, have the form

$$
\alpha a^{f(x)}+\beta b^{f(x)}+\gamma a^{f(x)} \geqslant 0
$$

or

$$
\alpha a^{f(x)}+\beta b^{f(x)}+\gamma c^{f(x)} \leqslant 0
$$

where $\alpha \neq 0, \beta, \gamma$ are real numbers, $f(x)$ is some function, and the bases $a, b, c$ satisfy the condition $b^{2}=a c$.