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

Sep 1, 2016

$x = 1 + \sqrt{1 + e}$

#### Explanation:

Start by applying the rule ${\log}_{a} \left(n\right) + {\log}_{a} \left(m\right) = {\log}_{a} \left(n \times m\right)$.

$\ln \left(x \left(x - 2\right)\right) = 1$

$\ln \left({x}^{2} - 2 x\right) = 1$

${x}^{2} - 2 x = {e}^{1}$

${x}^{2} - 2 x = e$

${x}^{2} - 2 x - e = 0$

$x = \frac{- \left(- 2\right) \pm \sqrt{- {2}^{2} - \left(4 \times 1 \times - e\right)}}{2 \times 1}$

$x = \frac{2 \pm \sqrt{4 + 4 e}}{2}$

$x = \frac{2 \pm \sqrt{4 \left(1 + e\right)}}{2}$

$x = \frac{2 \pm 2 \sqrt{1 + e}}{2}$

$x = 1 \pm \sqrt{1 + e}$

However, the $-$ solution is extraneous, since it renders the original equation undefined. So, the only actual solution is $x = 1 + \sqrt{1 + e}$

Hopefully this helps!