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

Oct 17, 2015

$1 + \sqrt{1 + e} \approx 2.9283$

#### Explanation:

Use the properties of logarithms to rewrite the expression.

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

Rewrite both sides in terms of the base $e$

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

From the properties of logarithms this becomes

$x \left(x - 2\right) = e$

Distribute $x$ on left hand side

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

Subtract $e$ form both sides

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

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

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

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

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

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

We are only interested in $1 + \sqrt{1 + e} \approx 2.9283$

The other root is negative and not in the domain of the logarithmic function.