7.200. $\frac{1+2 \log _{9} 2}{\log _{9} x}-1=2 \log _{x} 3 \cdot \log _{9}(12-x)$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}0<x<12, \\ x \neq 1 .\end{array}\right.$

Switch to base 3. Then we get

$$
\begin{aligned}
& \frac{1+\frac{2 \log _{3} 2}{\log _{3} 9}}{\frac{\log _{3} x}{\log _{3} 9}}-1=\frac{2}{\log _{3} x} \cdot \frac{\log _{3}(12-x)}{\log _{3} 9} \Leftrightarrow \frac{2+2 \log _{3} 2}{\log _{3} x}-1=\frac{\log _{3}(12-x)}{\log _{3} x} \Leftrightarrow \\
& \Leftrightarrow \frac{2+2 \log _{3} 2-\log _{3} x}{\log _{3} x}=\frac{\log _{3}(12-x)}{\log _{3} x} \Leftrightarrow 2+2 \log _{3} 2-\log _{3} x=\log _{3}(12-x) \Leftrightarrow \\
& \Leftrightarrow 2+2 \log _{3} 2=\log _{3} x+\log _{3}(12-x) \Leftrightarrow \\
& \Leftrightarrow \log _{3} 9+\log _{3} 4=\log _{3} x+\log _{3}(12-x), \log _{3} 36=\log _{3} x(12-x)
\end{aligned}
$$

$$
\text { hence } 36=x(12-x) \text { or } x^{2}-12 x+36=0,(x-6)^{2}=0, x=6 \text {. }
$$

Answer: 6.