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

##### 1 Answer
Mar 2, 2016

I found (depending upon the base $b$ of your logs):
$x = \frac{b + 1}{b - 1}$

#### Explanation:

We do not know the base of the logs (it could be $10$...) so we say that the base is $b$:
we get:
${\log}_{b} \left(x + 1\right) - {\log}_{b} \left(x - 1\right) = 1$
we use the property of logs that tells us:
$\log x - \log y = \log \left(\frac{x}{y}\right)$
and write:
${\log}_{b} \left(\frac{x + 1}{x - 1}\right) = 1$
we now use the definition of log:
${\log}_{b} x = a \to x = {b}^{a}$
$\frac{x + 1}{x - 1} = {b}^{1}$
rearranging:
$x + 1 = b \left(x - 1\right)$
$x + 1 = b x - b$
$x - b x = - 1 - b$
$x \left(1 - b\right) = - 1 - b$
$x = \frac{- 1 - b}{1 - b} = - \frac{b + 1}{1 - b} = \frac{b + 1}{b - 1}$

Now you can choose the base $b$ of your original logs and find $x$.
If, for example, $b$ were $10$ then you get:
$x = \frac{11}{9}$