# How do you solve log_5 (x+3)= 3 + log_5 (x-3) ?

Apr 10, 2016

$x = \frac{189}{62} = 3 \frac{3}{62}$

#### Explanation:

As ${\log}_{b} a = \log \frac{a}{\log} b$

${\log}_{5} \left(x + 3\right) = 3 + {\log}_{5} \left(x - 3\right)$ can be written as

$\log \frac{x + 3}{\log} 5 = 3 + \log \frac{x - 3}{\log} 5$

As $\log 5 \ne 0$, multiplying both sides by $\log 5$

$\log \left(x + 3\right) = 3 \log 5 + \log \left(x - 3\right)$ or

$\left(x + 3\right) = {5}^{3} \left(x - 3\right) = 125 \left(x - 3\right)$ or

$x + 3 = 125 x - 375$ or

$124 x = 378$ and $x = \frac{378}{124} = \frac{189}{62} = 3 \frac{3}{62}$