# How do you solve log_3(x-6)=log_9(2x)?

Sep 30, 2016

$x = 7 + \sqrt{13}$

#### Explanation:

${\log}_{3} \left(x - 6\right) - {\log}_{9} \left(2 x\right) = 0$

$\log \frac{x - 6}{\log} 3 - \log \frac{2 x}{\log} 9 = 0$

$\log \frac{x - 6}{\log} 3 - \log \frac{2 x}{2 \log 3} = 0$

Put on a common denominator:

$\frac{2 \log \left(x - 6\right)}{2 \log 3} - \log \frac{2 x}{2 \log 3} = 0$

${\log}_{9} {\left(x - 6\right)}^{2} - {\log}_{9} \left(2 x\right) = 0$

${\log}_{9} \left(\frac{{x}^{2} - 12 x + 36}{2 x}\right) = 0$

$\frac{{x}^{2} - 12 x + 36}{2 x} = {9}^{0}$

$\frac{{x}^{2} - 12 x + 36}{2 x} = 1$

${x}^{2} - 12 x + 36 = 2 x$

${x}^{2} - 14 x + 36 = 0$

$1 \left({x}^{2} - 14 x + 49 - 49\right) = - 36$

$1 {\left(x - 7\right)}^{2} - 49 = - 36$

$1 {\left(x - 7\right)}^{2} = 13$

$\left(x - 7\right) = \pm \sqrt{13}$

$x = 7 \pm \sqrt{13}$

However, the $x = 7 - \sqrt{13}$ is extraneous, because it renders the initial equation undefined.

Hopefully this helps!