9.135. a) $\frac{3 \log _{a} x+6}{\log _{a}^{2} x+2}>1$; б) $\log _{2} \log _{4} x+\log _{4} \log _{2} x \leq-4$.

9.135. a) $\frac{3 \log _{a} x+6}{\log _{a}^{2} x+2}>1$; b) $\log _{2} \log _{4} x+\log _{4} \log _{2} x \leq-4$.
## Solution.

a) Rewrite the inequality as $\frac{\log _{a}^{2} x-3 \log _{a} x-4}{\log _{a}^{2} x+2} \leq 0$ for $0 < a < 1$, and $\frac{\log _{a}^{2} x-3 \log _{a} x-4}{\log _{a}^{2} x+2} \geq 0$ for $a > 1$. If $\log _{a}^{2} x+2 > 0$, then $\log _{a}^{2} x-3 \log _{a} x-4 \leq 0$ for $0 < a < 1$, and $\log _{a}^{2} x-3 \log _{a} x-4 \geq 0$ for $a > 1$.

b) $\log _{2} \log _{4} x+\log _{4} \log _{2} x \leq-4$. Converting to base 2, we have $\log _{2} \log _{2} \sqrt{x}+\log _{2} \sqrt{\log _{2} x} \leq-4 \Rightarrow \log _{2}\left(\log _{2} \sqrt{x} \cdot \sqrt{\log _{2} x}\right) \leq-4$, $\log _{2} \sqrt{x} \cdot \sqrt{\log _{2} x} \leq \frac{1}{16}, \log _{2} x \cdot \sqrt{\log _{2} x} \leq \frac{1}{8}$. Squaring both sides of the inequality, we get $\log _{2}^{3} x \leq \frac{1}{64}$. Therefore, $0<\log _{2} x \leq \frac{1}{4}, 1<x \leq \sqrt[4]{2}$.

Answer: $x \in(1 ; \sqrt[4]{2})$.