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

Nov 24, 2015

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

#### Explanation:

Before beginning, note that we do not accept values less than $0$ for the $\ln$ function, and thus as we are beginning with $\ln \left(x\right)$ in the equation, it must be that $x > 0$.

We will use the following properties of logarithms:
(1) $\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$
(2) ${e}^{\ln \left(a\right)} = a$

as well as the quadratic formula:
$a {x}^{2} + b x + c = 0 \implies x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

By (1)
$\ln \left(x\right) + \ln \left(x + 1\right) = \ln \left(x \left(x + 1\right)\right) = \ln \left({x}^{2} + x\right)$
$\implies \ln \left({x}^{2} + x\right) = 2$

Then, by (2)
${e}^{2} = {e}^{\ln \left({x}^{2} + x\right)} = {x}^{2} + x$
$\implies {x}^{2} + x - {e}^{2} = 0$

Now we apply the quadratic formula with $a = b = 1$ and $c = - {e}^{2}$

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

But as noted at the beginning, we must have $x > 0$ and so we reject $\frac{- 1 - \sqrt{1 + {e}^{2}}}{2}$

Thus we are left with the answer

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