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

Jan 25, 2016

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

#### Explanation:

We can simplify the left hand side by using the logarithmic rule: $\ln \left(a\right) + \ln \left(b\right) = \ln \left(a b\right)$

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

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

To undo the natural logarithm, exponentiate both sides with base $e$.

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

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

Move the $e$ to the left side and solve with the quadratic equation.

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

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

Factor a $4$ from inside the square root, which can be pulled out as a $2$.

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

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

Notice that the negative root has been taken away, since for $\ln \left(a\right) , a > 0$ (if we are forgoing complex roots).