Example 16. Solve the inequality

$$
6 \sqrt[x]{9}-13 \sqrt[x]{3} \sqrt[x]{2}+6 \sqrt[x]{4} \leqslant 0
$$
Solution. The domain of admissible values of this inequality consists of all natural numbers greater than 1.

Inequality (12) is homogeneous. Dividing both parts by $\sqrt[x]{9}$ and setting $t=\sqrt[x]{2 / 3}$, we get

$$
6-13 t+6 t^{2} \leqslant 0
$$

The solution to this inequality is the interval $2 / 3 \leqslant t \leqslant 3 / 2$. Thus, inequality (12) is equivalent to the double inequality

$$
2 / 3 \leqslant \sqrt[x]{2 / 3} \leqslant 3 / 2
$$

from which we find

$$
\log _{2 / 3} \frac{3}{2} \leqslant \frac{1}{x} \leqslant \log _{2 / 3} \frac{2}{3}
$$

i.e., inequality (12) is satisfied by all numbers from the domain of admissible values. Therefore, the solution to inequality (12) is the set $X=\{x: x \geqslant 2, x \in \mathbf{N}\}$.

Some exponential inequalities contain expressions of the form $f(x)^{g(x)}$. Let us recall that by definition

$$
f(x)^{g(x)}=a^{g(x)} \log a f(x), \quad a>0, a \neq 1
$$

i.e., the function $f(x) g^{(x)}$ is defined when both functions $f(x), g(x)$ are defined and, in addition, $f(x)>0$.