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

Mar 19, 2018

$x = \frac{1}{2} + \frac{1}{2} \sqrt{1 + 4 {e}^{2}}$

#### Explanation:

We have:

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

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

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

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

$\therefore {\left(x - \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} - {e}^{2} = 0$

$\therefore {\left(x - \frac{1}{2}\right)}^{2} = \frac{1}{4} + {e}^{2} = 0$

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

$\therefore x - \frac{1}{2} = \pm \frac{1}{2} \sqrt{1 + 4 {e}^{2}}$

$\therefore x = \frac{1}{2} \pm \frac{1}{2} \sqrt{1 + 4 {e}^{2}}$

Strictly speaking we require that both:

$x > 0$ and $x + 1 > 0$

So that the logarithms, in the original equation exist. Consequently we can negate one solution and we have:

$x = \frac{1}{2} + \frac{1}{2} \sqrt{1 + 4 {e}^{2}}$