# How do you solve log(x-3)+log(x+5)=log9?

Jan 4, 2017

$x = 4$, and possibly $x = - 6$, depending upon the context of the question and what led to the logarithmic equarion

#### Explanation:

$\log \left(x - 3\right) + \log \left(x + 5\right) = \log 9$

Using $\log A B = \log A + \log B$ we get:

$\log \left(x - 3\right) \left(x + 5\right) = \log 9$

Using $\log A = \log B \iff A = B$ we have

$\left(x - 3\right) \left(x + 5\right) = 9$
$\therefore {x}^{2} + 5 x - 3 x - 15 = 9$
$\therefore {x}^{2} + 2 x - 24 = 0$
$\therefore \left(x + 6\right) \left(x - 4\right) = 0$
$\therefore x = 4 , - 6$

Now depending upon the context of the question we could probably eliminate $x = - 6$.

$x = - 6$, in the original form of the question, would lead to the log of negative numbers which do not exist.