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

Jul 10, 2015

I found: $x = - \frac{1 + e}{1 - e}$

#### Explanation:

I would rearrange your equation as:
$\ln \left(x + 1\right) - \ln \left(x - 1\right) = 1$
now I use the fact that:
$\log a - \log b = \log \left(\frac{a}{b}\right)$
to get:
$\ln \left(\frac{x + 1}{x - 1}\right) = 1$
we can take the exponential of both sides:
${e}^{\ln \left(\frac{x + 1}{x - 1}\right)} = {e}^{1}$ eliminating the $\ln$;
$\frac{x + 1}{x - 1} = e$
$x + 1 = e x - e$
rearranging:
$x - e x = - 1 - e$
$x \left(1 - e\right) = - \left(1 + e\right)$
$x = - \frac{1 + e}{1 - e}$