Re-arranging exponential equation Not quite applicable to many situations, but I'm extending this as more of a "I need help/tips" on re-arranging an equation.

I have something like this:

$$
0.02=\frac{1}{(1+c)}\left(1-e^{-\frac{Z}{y/(1+c))}}\right)
$$
I want to re-arranging this equation for $c$. How would this be done? I know that this is perhaps not the sort of question that should be asked here, but it seems to me a little tricky than I first thought, and I'm all out of ideas!

Am I looking at having to perform a numerical solution?
These equations need the Lambert W function, defined by $W(xe^x)=x$.
Equivalently, if $y=x+e^x$, then $e^y=e^xe^{e^x}$, so $e^x=W(e^y)$, and $x=\log W(e^y)$.
Let $v=\frac{-z(1+c)}y$, then the equation becomes 
$$\frac {-yv}{50z}+e^v=1\\
v+\log(-50z/y)+e^{v+\log(-50z/y)}=-50(z/y)+\log(-50z/y)\\
\text{Let }u=v+\log(-50z/y)\\
u+e^u=-50(z/y)+\log(-50z/y)\\
u=\log W(-50(z/y)e^{-50z/y})$$
from which you can untangle $c$.
These equations need the Lambert W function, defined by $W(xe^x)=x$.
Equivalently, if $y=x+e^x$, then $e^y=e^xe^{e^x}$, so $e^x=W(e^y)$, and $x=\log W(e^y)$.
Let $v=\frac{-z(1+c)}y$, then the equation becomes 
$$\frac {-yv}{50z}+e^v=1\\
v+\log(-50z/y)+e^{v+\log(-50z/y)}=-50(z/y)+\log(-50z/y)\\
\text{Let }u=v+\log(-50z/y)\\
u+e^u=-50(z/y)+\log(-50z/y)\\
u=\log W(-50(z/y)e^{-50z/y})$$
from which you can untangle $c$.