9.075. $\log _{2}\left(1+\log _{\frac{1}{9}} x-\log _{9} x\right)<1$.
Solution.

The given inequality is equivalent to the double inequality

$0<1+\log _{\frac{1}{9}} x-\log _{9} x<2, -1<\log _{\frac{1}{9}} x-\log _{9} x<1$.

We will change the base to $\frac{1}{9}$. We have

$-1<\log _{\frac{1}{9}} x+\log _{\frac{1}{9}} x<1, -1<2 \log _{\frac{1}{9}} x<1, -\frac{1}{2}<\log _{\frac{1}{9}} x<\frac{1}{2} \Leftrightarrow$

$\Leftrightarrow \frac{1}{3}<x<3$.

Answer: $\quad x \in\left(\frac{1}{3} ; 3\right)$