# How do you solve ln x + ln (x-2) = 1?

May 20, 2016

$x = 1 + \sqrt{1 + e}$
From ${\log}_{e} x + {\log}_{e} \left(x - 2\right) = {\log}_{e} \left(e\right)$ we conclude
${\log}_{e} \left(x \left(x - 2\right)\right) = {\log}_{e} \left(e\right)$ and also
$x \left(x - 2\right) = e$. Solving for $x$ we obtain
$x = 1 + \sqrt{1 + e}$