# How do you solve ln(2x-5) - ln(4x)= 2?

Jul 27, 2015

I found: x=5/(2(1-2e^2). But it cannot be accepted.

#### Explanation:

I would use one of the rules of logs as:
${\log}_{a} x - {\log}_{a} y = {\log}_{a} \left(\frac{x}{y}\right)$ to get:
$\ln \left(\frac{2 x - 5}{4 x}\right) = 2$ solving the log (with $\ln = {\log}_{e}$) you get:
$\frac{2 x - 5}{4 x} = {e}^{2}$
rearranging:
$2 x - 5 = 4 x {e}^{2}$
$2 x - 4 x {e}^{2} = 5$
$2 x \left(1 - 2 {e}^{2}\right) = 5$
so that:
$x = \frac{5}{2 \left(1 - 2 {e}^{2}\right)} = - 0.363$
BUT it is negative!
If you substitute back the arguments of the $\ln$ become negative!!!
We cannot accept them; so NO (real) SOLUTIONS!