Example 10. Solve the inequality

$$
\log _{4}\left(3^{x}-1\right) \log _{1 / 4} \frac{3^{x}-1}{16} \leq \frac{3}{4}
$$
Solution. Let $y=3^{x}-1$; then the given inequality takes the form

$$
\log _{4} y \log _{1 / 4} \frac{y}{16} \leqslant \frac{3}{4}
$$

Since

$$
\log _{1 / 4} \frac{y}{16}=-\log _{4} \frac{y}{16}=-\left(\log _{4} y-\log _{4} 16\right)=2-\log _{4} y
$$

the inequality (4) can be rewritten as

$$
2 \log _{4} y-\log _{4}^{2} y \leqslant 3 / 4
$$

Letting $t=\log _{4} y$, we have the inequality

$$
t^{2}-2 t+3 / 4 \geqslant 0
$$

the solutions of which are the intervals $-\infty<t \leqslant 1 / 2$ and $3 / 2 \leqslant t<+\infty$.

Thus, to find the values of $y$, we have the system of two simple logarithmic inequalities

$$
\left[\begin{array}{l}
\log _{4} y \leqslant 1 / 2 \\
\log _{4} y \geqslant 3 / 2
\end{array}\right.
$$

The solution to this system is the intervals $0<y \leqslant 2$ and $8 \leqslant y<+\infty$.

Therefore, the original inequality is equivalent to the system of two exponential inequalities

$$
\left[\begin{array}{r}
0<3^{x}-1 \leqslant 2 \\
3^{x}-1 \geqslant 8
\end{array}\right.
$$

i.e., the system

$$
\left[\begin{array}{r}
1<3^{x} \leqslant 3 \\
3^{x} \geqslant 9
\end{array}\right.
$$

The solution to the first inequality in this system is the interval $0<x \leqslant 1$, and the solution to the second is the interval $2 \leqslant x<+\infty$.

Thus, the original inequality holds for all values of $x$ from the intervals $0<x \leqslant 1$ and $2 \leqslant x<+\infty$.

There are various ways to present the solution to a logarithmic inequality. The most common are the method of transition to the solution of equivalent systems of inequalities and the method of partitioning the domain of the given inequality into intervals, on which the corresponding equivalent (on the considered interval) inequalities are solved. In essence, these methods of solving are the same and differ only in the way of presentation.