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

Aug 29, 2015

$x = \frac{1 + \sqrt{1 + 4 e}}{2} = \text{(approx.)} 2.22287$

#### Explanation:

If
$\textcolor{w h i t e}{\text{XXX}} \ln \left(x\right) + \ln \left(x - 1\right) = 1$
then
$\textcolor{w h i t e}{\text{XXX}} {e}^{\ln \left(x\right) + \ln \left(x - 1\right)} = {e}^{1}$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} {e}^{\ln \left(x\right)} \cdot {e}^{\ln \left(x - 1\right)} = {e}^{1}$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} x \cdot \left(x - 1\right) = e$

$\rightarrow \textcolor{w h i t e}{\text{XXX}} {x}^{2} - x - e = 0$

$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{1 \pm \sqrt{1 + 4 e}}{2}$
Since $\left(\sqrt{1 + 4 e} > 1\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} \frac{1 - \sqrt{1 + 4 e}}{2}$ is negative
Since $\ln \left(x\right)$ is undefined for $x < 0$
$\textcolor{w h i t e}{\text{XXX}} x = \frac{1 - \sqrt{1 + 4 e}}{2}$ is an extraneous solution
$\textcolor{w h i t e}{\text{XXX}} x = \frac{1 + \sqrt{1 + 4 e}}{2}$ is valid
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