# 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}

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

Now, $x=9+2^{1+\sqrt{4-\log_26}}$ or $x=9+2^{1-\sqrt{4-\log_26}}$