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

Jul 10, 2016

$x = \frac{e + 1}{e - 1}$
${\log}_{e} \left(x + 1\right) - 1 = {\log}_{e} \left(x - 1\right) \to {\log}_{e} \left(\frac{x + 1}{e}\right) = \log {e}_{e} \left(x - 1\right) \to \frac{x + 1}{e} = \left(x - 1\right)$
$x = \frac{e + 1}{e - 1} > 1 \to$feasible solution