Thread: e^x + x = 1

1. e^x + x = 1

Hi everyone,
This is my first post in this forum so please forgive me if I haven't posted this question in the right place.

The problem is that I can't figure out how to solve for x in the following....

e^x + x = 1

The usual index and log rules don't seem to help much.

Any ideas?

Thanks.

2. This equation has only one root x=0. Proof that. Use derivatives for example

3. Originally Posted by ICanFly
This equation has only one root x=0. Proof that. Use derivatives for example
There actually are infinitely many complex roots, $\displaystyle x = 1 - W_n(e)$ with $\displaystyle n \in \mathbb{Z}$ (along with the real root $\displaystyle x = 0$), but perhaps you don't want to mess with this for the moment ... ($\displaystyle W_n$ is the analytic continuation of the logarithm function).

4. Originally Posted by MrRecursive
Hi everyone,
This is my first post in this forum so please forgive me if I haven't posted this question in the right place.

The problem is that I can't figure out how to solve for x in the following....

e^x + x = 1

The usual index and log rules don't seem to help much.

Any ideas?

Thanks.
This is equivalent to the function

$\displaystyle y = e^x + x - 1 = 0$.

By inspection, it's easy to see that $\displaystyle x = 0$ satisfies this equation.

Now, by Rolle's Theorem, if a function has more than one root, it has to turn around. So if we can show that the function does not have any turning points, then it can't have more than one root.

$\displaystyle \frac{dy}{dx} = e^x + 1$

$\displaystyle 0 = e^x + 1$

$\displaystyle e^x = -1$

But since $\displaystyle e^x > 0$ for all $\displaystyle x$, this equation is not solvable (at least in the reals).

So there are not any turning points, which means there is only one root.

So $\displaystyle x = 0$ is the only root.