Example 18. Solve the inequality

$$
\log _{x} 2 x \leqslant \sqrt{\log _{x}\left(2 x^{3}\right)}
$$
Solution. The given inequality is equivalent to the inequality

$$
\log _{x} 2+10$, $x \neq 1$.

If $0 < x < 1$, then we have

$$
\left\{\begin{array} { l } 
{ 1 / \operatorname { l o g } _ { 2 } x \geqslant 1 } \\
{ 1 / \operatorname { l o g } _ { 2 } x \leqslant - 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\log _{2} x \leqslant 1 \\
\log _{2} x \leqslant-1 / 3
\end{array} \Leftrightarrow \log _{2} x \leqslant -1 / 3 \Leftrightarrow x \leqslant 1 / \sqrt[3]{2}\right.\right.
$$

Thus, on the set $0 < x < 1$, the solution to the system of inequalities (13) is the interval $0 < x \leqslant 1 / \sqrt[3]{2}$.

If $x > 1$, then we have

$$
\left\{\begin{array} { l } 
{ 1 / \operatorname { l o g } _ { 2 } x \leqslant 1 } \\
{ 1 / \operatorname { l o g } _ { 2 } x \geqslant - 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\log _{2} x \geqslant 1 \\
\log _{2} x \geqslant-1 / 3
\end{array} \Leftrightarrow \log _{2} x \geqslant 1 \Leftrightarrow x \geqslant 2\right.\right.
$$

Thus, on the set $x > 1$, the solution to the system of inequalities (13) is the interval $2 \leqslant x < +\infty$.

Therefore, the set of all solutions to the system (13), and hence to the original inequality, consists of the intervals $0 < x \leqslant 1 / \sqrt[3]{2}$ and $2 \leqslant x < +\infty$.

## 194 CH. 3. INEQUALITIES WITH ONE UNKNOWN