<s>[INST] How to solve the equation $\log_2(x-9)+\log_{(2x-18)}6=3$. 
Solve the following equation. $$\log_2(x-9)+\log_{(2x-18)}6=3.$$ 

I tried this way, 
\begin{align}
  \log_2(x-9)+\log_{(2x-18)}6 & \ = \ 3\\
  \log_2(x-9)+\log_{2(x-9)}6 & \ = \ 3\\
  \log_2(x-9)+\frac{\log_{2}6}{\log_2(2(x-9))} & \ = \ 3\\
  \log_2(x-9)+\frac{\log_{2}6}{1+\log_2(x-9)} & \ = \ 3\\
  u+\frac{\log_{2}6}{1+u} & \ = \ 3&&\text{(when $ u=\log(x-9) $)}\\
  u(1+u)+\log_{2}6 & \ = \ 3(1+u)\\
  u^2-2u+(\log_{2}6-3) & \ = \ 0\\
  u & \ = \ \frac{2\pm\sqrt{-4\log_26+16}}{2}\\
  & \ = \ 1\pm\sqrt{-2\log_26+8}
\end{align}
 [/INST] Now, $x=9+2^{1+\sqrt{4-\log_26}}$ or $x=9+2^{1-\sqrt{4-\log_26}}$
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