Example 14. Solve the inequality

$$
10^{7 x-1}+6 \cdot 10^{1-7 x}-5 \leqslant 0
$$
Solution. Using the properties of exponents, rewrite the given inequality as

$$
10^{7 x-1}+\frac{6}{10^{7 x-1}}-5 \leq 0
$$

Let $t=10^{7 x-1}$, we get

$$
\left\{\begin{array} { l } 
{ t > 0 , } \\
{ t + 6 / t - 5 \leqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
t>0, \\
t^{2}-5 t+6 \leqslant 0
\end{array} \Leftrightarrow 2 \leqslant t \leqslant 3\right.\right.
$$

Thus, the given inequality is equivalent to the double inequality

$$
2 \leqslant 10^{7 x-1} \leqslant 3
$$

from which we find

$$
\frac{1}{7}(1+\lg 2) \leqslant x \leqslant \frac{1}{7}(1+\lg 3)
$$

Therefore, the interval $[(1+\lg 2) / 7 ;(1+\lg 3) / 7]$ is the set of all solutions to the given inequality.

Solving some exponential inequalities reduces to solving algebraic homogeneous inequalities, i.e., inequalities of the form

$a_{0} f^{n}(x)+a_{1} f^{n-1}(x) g(x)+a_{2} f^{n-2}(x) g^{2}(x)+\ldots$

$$
\ldots+a_{n-1} f(x) g^{n-1}(x)+a_{n} g^{n}(x) \geqslant 0
$$

where $a_{0}, a_{1}, \ldots, a_{n}$ are some numbers $\left(a_{0} \neq 0\right)$, and $f(x), g(x)$ are some functions.