# How do you solve  log (2x + 2) – log (x + 6) =0?

Mar 7, 2018

$x = 4$

#### Explanation:

Note 1:
$\textcolor{w h i t e}{\text{XXX}} \log \left(A\right) - \log \left(B\right) = \log \left(\frac{A}{B}\right)$
Note 2:
$\textcolor{w h i t e}{\text{XXX}} \log \left(1\right) = 0$

Therefore
$\textcolor{w h i t e}{\text{XXX}} \log \left(2 x + 2\right) - \log \left(x + 6\right) = 0$
can be written as
$\textcolor{w h i t e}{\text{XXX}} \log \left(\frac{2 x + 2}{x + 6}\right) = \log \left(1\right)$
which implies
$\textcolor{w h i t e}{\text{XXX}} \frac{2 x + 2}{x + 6} = 1$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow 2 x + 2 = x + 6$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow x = 4$