How do I prove this transcendental equation has a solution?

Question

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

Answer 1

Edit: The question was changed to eliminate the solution $B=0$.

First answer: By inspection, $B=0$ is a solution. In the case $c=1$, it is the only solution.

Answer to modified question: Consider the function $f(x)=1-x-e^{-cx}$. We have $f'(x)=ce^{-cx}-1$. This is positive at $x=0$, since $c\gt 1$. Thus by continuity $f'(x)\gt 0$ for a while after $x=0$.

It follows that $f(x)$ is increasing for a while past $x=0$, and in particular is positive for some $a\gt 0$.

When $x$ is large enough, $f(x)$ is negative, since $\lim_{x\to\infty}e^{-cx}=0$.

Thus by the Intermediate Value Theorem, there is an $x\gt 0$ such that $f(x)=0$.

Remark: By closer examination of the derivative, one can show that there is exactly one $x\gt 0$ such that $f(x)=0$.

Answer 2

Note that for $c \gt 1$, for $B$ slightly greater than zero $1-B \gt e^{-cB}$ (you can use the derivatives to show this). At $B=1$ we have $1-B \lt e^{-cB}$ and there must be a point where they cross.

Answer 3

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 $W$.