# How do you solve ln (x-3) - ln (x+4)=ln(x-1) - ln(x+2)?

Nov 20, 2015

I found: NO REAL $x$.

#### Explanation:

You can start by using the property of logs that tells us:
$\log a - \log b = \log \left(\frac{a}{b}\right)$
and write:
$\ln \left(\frac{x - 3}{x + 4}\right) = \ln \left(\frac{x - 1}{x + 2}\right)$
if the two logs must be equal their arguments must be as well; so:
$\frac{x - 3}{x + 4} = \frac{x - 1}{x + 2}$ rearranging and getting rid of the denominators:
$\left(x - 3\right) \left(x + 2\right) = \left(x - 1\right) \left(x + 4\right)$
$\cancel{{x}^{2}} + 2 x - 3 x - 6 = \cancel{{x}^{2}} + 4 x - x - 4$
isolate $x$:
$- x - 3 x = 6 - 4$
$- 4 x = 2$
$x = - \frac{2}{4} = - \frac{1}{2}$
If you plug this solution back into the original equation, you get some negative arguments of logs which prompt us to exclude $x = - \frac{1}{2}$.