# How do you solve ln x + ln (x-3) =0?

Jul 29, 2016

$x = \frac{3 + \sqrt{13}}{2}$

#### Explanation:

${\log}_{e} x + {\log}_{e} \left(x - 3\right) = {\log}_{e} 1$ so
${\log}_{e} x \left(x - 3\right) = {\log}_{e} 1$

or

$x \left(x - 3\right) = 1 \to {x}^{2} - 3 x - 1 = 0$

with the solutions

$x = \frac{3 \pm \sqrt{13}}{2}$ now, cheking the feasibility we choose

$x = \frac{3 + \sqrt{13}}{2}$ because with this value,

$x > 0$ and $x - 3 > 0$