How do I prove this transcendental equation has a solution? I am trying to prove that for the following equation, there is a B that solves it (c is a constant):
$1-B = e^{-cB}$
I understand this is a transcendental equation, but how do I prove there is a B that solves it?
I need a non-zero solution. c > 1

I'm going to use $x$ for $B$:
$1-x = e^{-cx}$
http://www.wolframalpha.com/input/?i=1-x+%3D+e%5E%28-cx%29 solves for x as


*

*1 $$x = \frac {W(-c*e^{-c} )+c} {c}$$ where W is the product log function. 


To prove this, we need to show that $$(cx-c)e^{cx-c}+c*e^{-c}=0$$
Let's use an alternate form of the original equality(this is trivial):


*

*2 $$(x-1)e^{cx}=-1$$


Assume $(cx-c)e^{cx-c}+c*e^{-c}=y$ 
$$(cx-c)e^{cx-c}=y-c*e^{-c}$$
Rewrite this as $$\frac {(cx-c)e^{cx}} {e^c}=y-c*e^{-c}$$
Now, from 2 we have $$(cx-c)e^{cx}=-c$$
Substituting that gives $$-\frac {c} {e^c}=y-c*e^{-c}$$
which trivially leads to $y=0$, which then leads to 1 above, using the definition  of $W$.
