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

Jun 26, 2016

Do some algebra and apply some log properties to get $x = - \frac{e + 1}{1 - e} \approx 2.164$

#### Explanation:

Begin by collecting the $x$ terms on one side of the equation and the constant terms (numbers) on the other:
$\ln \left(x + 1\right) - \ln \left(x - 1\right) = 1$

Apply the natural log property $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$:
$\ln \left(\frac{x + 1}{x - 1}\right) = 1$

Raise both sides to the power of $e$:
${e}^{\ln} \left(\frac{x + 1}{x - 1}\right) = {e}^{1}$
$\to \frac{x + 1}{x - 1} = e$

This equation is solved in a few steps:
$x + 1 = e \left(x - 1\right)$
$x + 1 = e x - e$
$x - e x = - e - 1$
$x \left(1 - e\right) = - e - 1$
$x = - \frac{e + 1}{1 - e} \approx 2.164$

Note that $\frac{- e - 1}{1 - e} = \frac{- 1 \left(e + 1\right)}{1 - e} = - \frac{e + 1}{1 - e}$.