# How do you solve ln (x + 2) + ln (x - 2) = 0?

Apr 26, 2016

Take exponents of both sides to get a quadratic, the positive root of which is the solution:

$x = \sqrt{5}$

#### Explanation:

Given:

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

If we are dealing with Real logarithms then taking exponents of both sides we find:

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

That is:

${x}^{2} - 4 = 1$

So:

${x}^{2} = 5$

Hence:

$x = \pm \sqrt{5}$

We can discard the negative root since we require $x + 2 > 0$ and $x - 2 > 0$.

So the remaining solution is $x = \sqrt{5}$

$\textcolor{w h i t e}{}$
Footnote

$x = - \sqrt{5}$ is not a solution even if we consider Complex logarithms.

If $t < 0$ then the principal Complex natural logarithm has the value:

$\ln t = \ln \left\mid t \right\mid + \pi i$

[[ More generally $\ln z = \ln \left\mid z \right\mid + A r g \left(z\right) i$ ]]

So we find:

$\ln \left(- \sqrt{5} + 2\right) + \ln \left(- \sqrt{5} - 2\right)$

$= \ln \left(\sqrt{5} - 2\right) + \pi i + \ln \left(\sqrt{5} + 2\right) + \pi i = 2 \pi i$