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

Jul 10, 2016

$\ln x - 2 \ln \left(x - 6\right) = 0$

$\ln x - \ln {\left(x - 6\right)}^{2} = 0$

$\ln x - \ln \left({x}^{2} - 12 x + 36\right) = 0$

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

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

$x = 1 \left({x}^{2} - 12 x + 36\right)$

$0 = {x}^{2} - 13 x + 36$

$0 = \left(x - 9\right) \left(x - 4\right)$

$x = 9 \mathmr{and} 4$

Checking the solutions in the original equation, we find that only $x = 9$ works. Hence, the solution set is $\left\{9\right\}$.

Hopefully this helps!