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

Nov 14, 2015

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

#### Explanation:

Using the rules of logarithms,

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

Therefore,

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

Then, we exponentiate both sides (put both sides to the $e$ power):

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

Simplify, remembering that exponents undo logarithms:

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

Now, we complete the square:

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

Simplify:

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

Take the square root of both sides:

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

Add $\frac{1}{2}$ to both sides:

x=(1±sqrt(4e+1))/2

Eliminate the negative answer (in ${\log}_{\text{a}} b , b > 0$):

$\implies \textcolor{b l u e}{x = \frac{1 + \sqrt{4e+1}}{2}}$