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

May 5, 2016

I found: $x = 1 + \sqrt{1 + e}$

#### Explanation:

I would write it as one log (multiplying the arguments) as:
$\ln \left[x \left(x - 2\right)\right] = 1$
then use the definition of log and write:
$x \left(x - 2\right) = {e}^{1}$
${x}^{2} - 2 x - e = 0$
use the Quadratic Formula to find $x$:
${x}_{1 , 2} = \frac{2 \pm \sqrt{4 + 4 e}}{2} = \frac{2 \pm 2 \sqrt{1 + e}}{2} = 1 \pm \sqrt{1 + e}$
I would use the positive one only, i.e.:
$x = 1 + \sqrt{1 + e}$