solve ln(x+2)+ln(x-1)=1
it can be written as (x+2)(x-1)=e apparently
What rule is this?
and the answer is x = (sqrt[9+4e]-1)/2
$\displaystyle (x+2)(x-1) = e$
$\displaystyle x^2 + x - 2 - e = 0$
$\displaystyle x^2 + x - (2+e) = 0$
$\displaystyle a = 1$, $\displaystyle b = 1$, $\displaystyle c = -(2+e)$
$\displaystyle x = \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot -(2+e)}}{2}$
$\displaystyle x = \frac{-1 \pm \sqrt{9+4e}}{2}$
since $\displaystyle x > 1$ ...
$\displaystyle x = \frac{-1 + \sqrt{9+4e}}{2}$