# How to solve for x: $e*x + e^{-x} = 0$

I already know the answer is supposed to be $x=-1$, but I have no idea how to get to that. What I've done so far is:

$ln(e^{-x}) = ln(e)+ln(-x)$
$-x = 1 + ln(-x)$
$ln(-x) + x = -1$ or $ln(x * -e^x) = -1$

I just don't know what to do from here. What can I use to now conclude that indeed $x=-1$?

• Knowing that $x=-1$ is a solution, you need to prove that it is unique. If you want to solve an equation of the form $ex+ae^{-x}=0$, you can use numerical methods. – Paracosmiste Nov 9 '14 at 20:24
• I wonder if there is a way to answer this without considering the derivative. – mick Nov 9 '14 at 20:39
• @mick There is if you define $e^x$ as $\sum_{n=0}^\infty\frac{x^n}{n!}$. – user2345215 Nov 9 '14 at 20:42
• @user2345215 yeah I posted that as comment a min ago. I gave +1 :) – mick Nov 9 '14 at 20:46

Let $f(x)= ex+e^{-x}$. Then $$f'(x)=e-e^{-x}$$
It is easy to prove that $f'(x) <0$ on $(-\infty,-1)$ and $f'(x) >0$ on $(-1, \infty)$. This implies that $f$ is strictly decreasing on $(- \infty, -1]$ and strictly increasing on $[-1, \infty)$.
Set $x={-y}-1$, then it becomes $$e^y=1+y$$ But $e^y\ge1+y$ with equality only when $y=0$ because $e^y$ is a convex function and $1+y$ is its tangent line at $y=0$.
• @mick It is, although it's not immediately obvious for negative $y$ (that's an easy excercise). – user2345215 Nov 9 '14 at 20:46
• @mick You can show $1+y\le 1+y+y^2/2!+y^3/3!+\ldots$ by noting it's trivial for $y\le-1$ (if you know $e^y$ is always positive) and for $y\in[-1,0]$, just group $y^{2k}/(2k)!+y^{2k+1}/(2k+1)!$ together to see it's positive. – user2345215 Nov 9 '14 at 20:50
Consider $f(x)=ex+e^{-x}$ then we have $f'(x)=e-e^{-x}$ and $f''(x)=e^{-x}>0$ thus the equation $f'(x)=0$ has only the solution $x=-1$ and this gives us a local minimum and $f(-1)=0$.