# How do you solve ln (x – 2) + ln (x + 2) = ln 5?

Sep 9, 2015

Use properties of $\ln$ to find $\ln \left({x}^{2} - 4\right) = \ln \left(5\right)$ and hence find solution: $x = 3$

#### Explanation:

$\ln \left({x}^{2} - 4\right) = \ln \left(\left(x - 2\right) \left(x + 2\right)\right) = \ln \left(x - 2\right) + \ln \left(x + 2\right)$

$= \ln \left(5\right)$

So ${x}^{2} - 4 = 5$

Add $4$ to both sides to get: ${x}^{2} = 9$

So $x = \pm 3$

If $x = - 3$, then $x - 2 < 0$ and $x + 2 < 0$, so neither $\ln \left(x - 2\right)$ nor $\ln \left(x + 2\right)$ have Real values.

If $x = 3$ then:

$\ln \left(x - 2\right) + \ln \left(x + 2\right) = \ln \left(1\right) + \ln \left(5\right) = 0 + \ln \left(5\right) = \ln \left(5\right)$

So the only solution is $x = 3$

Well you can write it as follows

$\ln \left(x - 2\right) + \ln \left(x + 2\right) = \ln 5 \implies \ln \left({x}^{2} - 4\right) = \ln 5 \implies {x}^{2} - 4 = 5 \implies {x}^{2} = 9 \implies x = - 3 , x = + 3$

But x-2>0 so the solution is x=3