9.183. $\log _{\frac{1}{2}} \log _{2} \log _{x-1} 9>0$.
Solution.

The original inequality is equivalent to the inequality $0<\log _{2} \log _{x-1} 9<1 \Leftrightarrow$ $\Leftrightarrow 1<\log _{x-1} 9<2$. We obtain two systems of inequalities:

1) $\left\{\begin{array}{l}0<x-1<1, \\ (x-1)^{2}<9<x-1,\end{array} \varnothing\right.$;

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0691.jpg?height=299&width=888&top_left_y=18&top_left_x=192)

Answer: $x \in(4 ; 10)$.