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

Feb 7, 2016

Since the logs are in the same base already, we can use the rule ${\log}_{a} b - {\log}_{a} c = {\log}_{a} \left(\frac{b}{c}\right)$

#### Explanation:

$\log \left(x + 1\right) - \log \left(x - 1\right) = 1$

$\log \left(\frac{x + 1}{x - 1}\right) = 1$

Since this is a base 10, we can now convert to exponential form.

$\left(\frac{x + 1}{x - 1}\right) = {10}^{1}$

$x + 1 = 10 \left(x - 1\right)$

$x + 1 = 10 x - 10$

$1 + 10 = 10 x - x$

$11 = 9 x$

$\frac{11}{9} = x$

Hopefully you understand now!