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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}

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  • $\begingroup$ You are almost done! Just consider $\log_2(x-9)=u=1\pm\sqrt{8-2\log_26}$ and solve it for $x$. $\endgroup$ – Mundron Schmidt Sep 2 '17 at 5:28
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Now, $x=9+2^{1+\sqrt{4-\log_26}}$ or $x=9+2^{1-\sqrt{4-\log_26}}$

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  • $\begingroup$ @Mundron Schmidt and Michael Rozenberg, Thanks for your attentions. $\endgroup$ – somkiat_t Sep 2 '17 at 5:40
  • $\begingroup$ @somkiat_t You are welcome! It was a mistake in your computations. I fixed your mistake in my solution. $\endgroup$ – Michael Rozenberg Sep 2 '17 at 5:43

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