# How do you solve ln(x) + ln(x-2) = ln(x+4)?

Oct 21, 2015

Use properties of logs to derive a quadratic equation in $x$, one of whose roots is a valid solution to the original equation.

$x = 4$

#### Explanation:

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

So

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

Subtract $x + 4$ from both ends to get:

$0 = {x}^{2} - 3 x - 4 = \left(x - 4\right) \left(x + 1\right)$

So $x = - 1$ or $x = 4$

Discard $x = - 1$ since both $\ln \left(- 1\right)$ and $\ln \left(- 1 - 2\right)$ are not defined.

So the only solution is $x = 4$