9.198. $9^{\log _{2}(x-1)-1}-8 \cdot 5^{\log _{2}(x-1)-2}>9^{\log _{2}(x-1)}-16 \cdot 5^{\log _{2}(x-1)-1}$.
## Solution.

Domain of definition: $x>1$.

$$
\begin{aligned}
& \text { We have } \frac{9^{\log _{2}(x-1)}}{9}-\frac{8 \cdot 5^{\log _{2}(x-1)}}{25}>9^{\log _{2}(x-1)}-\frac{16 \cdot 5^{\log _{2}(x-1)}}{5}, \\
& \frac{9^{\log _{2}(x-1)}}{9}-9^{\log _{2}(x-1)}>\frac{8 \cdot 5^{\log _{2}(x-1)}}{25}-\frac{16 \cdot 5^{\log _{2}(x-1)}}{5} \\
& -\frac{8}{9} \cdot 9^{\log _{2}(x-1)}>-\frac{72}{25} \cdot 5^{\log _{2}(x-1)}, \frac{1}{9} \cdot 9^{\log _{2}(x-1)}<\frac{9}{25} \cdot 5^{\log _{2}(x-1)} \\
& \frac{9^{\log _{2}(x-1)}}{9^{2}}<\frac{5^{\log _{2}(x-1)}}{5^{2}} \Leftrightarrow \frac{9^{\log _{2}(x-1)}}{5^{\log _{2}(x-1)}}<\frac{9^{2}}{5^{2}},\left(\frac{9}{5}\right)^{\log _{2}(x-1)}<\left(\frac{9}{5}\right)^{2} \Leftrightarrow \\
& \Leftrightarrow \log _{2}(x-1)<2,0<x-1<4,1<x<5
\end{aligned}
$$

Answer: $x \in(1 ; 5)$.