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

Sep 10, 2015

Use exponentiation to produce a quadratic equation to solve using the quadratic formula.

Discard the root of this quadratic that results in $x < 0$, leaving the solution:

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

#### Explanation:

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

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

Then using the quadratic formula, with $a = 1$, $b = - 2$ and $c = e$

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

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

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

Now $\sqrt{1 + e} > 1$, so $1 - \sqrt{1 + e} < 0$ and $\ln \left(1 - \sqrt{1 + e}\right)$ is not defined.

So the only valid solution is $x = 1 + \sqrt{1 + e}$