ln(x) + ln(x-1) = 1
Solve for x
First off, we can't have $\displaystyle x\leq 0$
Remember the rule
$\displaystyle \log_a(m)+\log_a(n)=\log_a(mn)$?
Using this rule we get
$\displaystyle \ln[x(x-1)]=1$
$\displaystyle x(x-1)=e^1$
$\displaystyle x(x-1)=e$
$\displaystyle x^2-x=e$
$\displaystyle x^2-x-e=0$
Then using the quadratic formula, namely
If $\displaystyle ax^2 + bx + c = 0$ then $\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
we get as our solution for x...
$\displaystyle x=\frac{1\pm \sqrt{(-1)^2-4\times 1 \times (-e)}}{2\times 1}$
$\displaystyle =\frac{1 \pm \sqrt{1+4e}}{2}$
And since we know $\displaystyle x > 0$ we disregard the negative answer.
So $\displaystyle x = \frac{1 + \sqrt{1 + 4e}}{2}$.