# How do you solve log _ 5 (x-3)+log _ 5(x+1) = log _ 5(x+3)?

Dec 13, 2015

$x = \frac{3 + \sqrt{33}}{2}$

#### Explanation:

Combine using logarithm rules.

${\log}_{5} \left(\left(x - 3\right) \left(x + 1\right)\right) = {\log}_{5} \left(x + 3\right)$

Raise both sides to the $5$th power.

${5}^{{\log}_{5} \left(\left(x - 3\right) \left(x + 1\right)\right)} = {5}^{{\log}_{5} \left(x + 3\right)}$

$\left(x - 3\right) \left(x + 1\right) = x + 3$

${x}^{2} - 2 x - 3 = x + 3$

${x}^{2} - 3 x - 6 = 0$

$x = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}$

Throw out the negative answer. It cannot be used because a logarithm has to be a function of a POSITIVE number.

Thus, $x = \frac{3 + \sqrt{33}}{2}$.