How to find $L$ if $L=\frac{c}{(1-L)^a}$ How to find $L$ if $L=\frac{c}{(1-L)^a}$
I was trying to apply log but $\ln L +a\ln (1-L)=\ln c$. How can continued please?
Thank you
 A: As Ethan Bolker notes, numerical methods are needed in general.  But the details work out rather nicely nevertheless.  We assume $a$ and $c$ are positive and define $$f(L) \doteq \ln L +a\ln (1-L) - \ln c.$$  The goal is to find all solutions to $f(L) = 0$ for $0 < L < 1$.  The function $f$ is smooth and concave, and $f(L) \rightarrow -\infty$ as $L \rightarrow 0$ and as $L \rightarrow 1$.  Therefore $f$ has a unique maximum on $(0,1)$ and it occurs at the solution to $$f'(L) = \frac{1}{L} - \frac{a}{1-L} = 0.$$  This solution is $L^* \doteq 1/(a+1)$, and the maximum that $f$ achieves there is $$f^* \doteq f(L^*) = a \ln a - (a+1) \ln(a+1) - \ln c.$$  We define $c^* \doteq a^a/(a+1)^{a+1}$ to be the value of $c$ for which this maximum is $0$.
For $c > c^*$ there are no solutions to  $f(L) = 0$.
For $c = c^*$ there is one solution, namely $L = L^*$.
And for $c < c^*$ there are exactly two solutions, one with $0 < L < L^*$ and one with $L^* < L < 1$.  To find these two solutions numerically, it is helpful to bound these roots a little better than with locations where $f(L) = -\infty$.  We can do this by solving $f(L) = 0$ but with the second or the first term of $f$ omitted.
This yields values of $L$ (on opposite sides of $L^*$) where $f(L) < 0$, so we obtain the intervals $(c,L^*)$ and $(L^*,1-c^{1/a})$ that contain each root in the $c < c^*$ case.  You may now confidently call a numerical root finder, telling it to find the unique root that exists in each of the given intervals.
