I need to solve the equation $$y=2500\ln{\left(\frac{70000}{70000-250x}\right)}-10x$$ for when $y=0$.
So far, I have simplified(?) the equation to give $$xe^{x/250}-280e^{x/250}+280=0$$ but at this point I am stuck. I know the two solutions are $x=0$ and $x\approx57.77$, although the latter I could only achieve by trial and error with a calculator, and I need to be able to show how I got these solutions.
Any pointers would be hugely appreciated!
Calling $y = \frac{x}{250}$
$$ 250 y e^y -280 e^y + 280 = 0 \Rightarrow (250y-280)e^y +280 = 0 $$
now making $z = 250y-280$
$$ ze^{\frac{z}{250}}e^{\frac{280}{250}}+280 = 0\Rightarrow 250\left(\frac{z}{250}e^{\frac{z}{250}}\right)+280 e^{-\frac{280}{250}} = 0 $$
now using the Lambert function
$\frac{z}{250} = W\left(-\frac{280}{250}e^{-\frac{280}{250}}\right)$
and finally
$$ \left\{ \begin{array}{rcl} x & = & 0\\ x & = & 10\left(28-25 W\left(-\frac{28}{25e^{28/25}}\right)\right)\approx 57.7766 \end{array} \right. $$