Example 4. Solve the inequality

$$
\log _{2} \frac{5-12 x}{12 x-8}+\log _{1 / 2} x \leqslant 0
$$
Solution. Since $\log _{1 / 2} x=-\log _{2} x$, the given inequality is equivalent to the inequality

$$
\log _{2} \frac{5-12 x}{12 x-8} \leqslant \log _{2} x
$$

## 182 CH. 3. INEQUALITIES WITH ONE UNKNOWN

which, taking into account that the base of the logarithm is greater than one, is equivalent to the system of inequalities

$$
\left\{\begin{array}{l}
\frac{5-12 x}{12 x-8}>0 \\
\frac{5-12 x}{12 x-8} \leq x
\end{array}\right.
$$

The inequality

$$
\frac{5-12 x}{12 x-8}>0 \Leftrightarrow \frac{x-5 / 12}{x-2 / 3}<0 \Leftrightarrow 5 / 12<x<2 / 3
$$

Since $12 x-8<0$ for the found values of $x$, on the interval $5 / 12<x<2 / 3$ the second inequality of the system (2) is equivalent to the inequality $5-12 x \geqslant x(12 x-8)$, from which $-5 / 6 \leqslant x \leqslant 1 / 2$. Of these values of $x$, only those that satisfy the double inequality $5 / 12<x<2 / 3$ are solutions to the original inequality.

Thus, the set of all solutions to the original inequality is the interval $5 / 12<x \leqslant 1 / 2$.