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

Oct 28, 2015

I found: $x = + \sqrt{11}$

#### Explanation:

You can use the rule of logs that relates the sum of logs of same base and the multiplication of the integrand as:
${\log}_{a} x + {\log}_{a} y = {\log}_{a} \left(x \cdot y\right)$
and get:

${\log}_{a} \left[\left(x + 1\right) \left(x - 1\right)\right] = 1$ using the definition of log:

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

Now it depends upon the base $a$ of your logs....and also if it the same for both the original ones!
If $a = 10$ then you have:
${x}^{2} - 1 = 10$
${x}^{2} = 11$
$x = \pm \sqrt{11}$
Where I accept the positive only, $x = + \sqrt{11}$

If $a$ isn't $10$ simply insert the right value into (1) and solve.