Example 12. Solve the inequality

$$
3 \cdot 7^{2 x}+37 \cdot 140^{x}<26 \cdot 20^{2 x}
$$
Solution. Let's write the inequality in the form

$$
3 \cdot 49^{x}+37 \cdot 140^{x}-26 \cdot 400^{x} \leqslant 0
$$

In this inequality, the numbers 49, 140, and 400 form three consecutive terms of a geometric progression with a common ratio of 20/7. Dividing both sides by $400^{x}$, we obtain the inequality

$$
3(7 / 20)^{2 x}+37(7 / 20)^{x}-26 \leq 0
$$

Let $t=(7 / 20)^{x}$, then we have

$$
\left\{\begin{array}{l}
t > 0, \\
3 t^{2} + 37 t - 26 \leq 0 \\
3(t+13)(t-2 / 3) < 0
\end{array} \Leftrightarrow 0 < t < 2 / 3\right.
$$

Thus, the inequality (9) is equivalent to the inequality $(7 / 20)^{x} \leqslant 2 / 3$, from which we get $x \geqslant \log _{7 / 20} \frac{2}{3}$. Therefore, the interval $\left[\log _{7 / 20} \frac{2}{3},+\infty\right)$ is the set of all solutions to the inequality (11).