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

Jul 28, 2016

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

#### Explanation:

We can change the sum of logs into a product of arguments as:
$\ln \left[x \left(x - 2\right)\right] = 1$
We then use the definition of log to get:
$x \left(x - 2\right) = {e}^{1}$
Rearranging:
${x}^{2} - 2 x - e = 0$
${x}_{1 , 2} = \frac{2 \pm \sqrt{4 + 4 e}}{2} = \frac{2 \pm \sqrt{4 \left(1 + e\right)}}{2} =$
$= \frac{2 \pm 2 \sqrt{1 + e}}{2} = \frac{2 \left(1 \pm \sqrt{1 + e}\right)}{2} = 1 \pm \sqrt{1 + e}$
${x}_{1} = 1 + \sqrt{1 + e}$
${x}_{2} = 1 - \sqrt{1 + e}$ NO, because negative.