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This seems to have stumped even my TA, so I'm asking it here.

Given $e^x + x = 1$, solve for $x$.

I already know that the answer is zero, but have no idea how to get there.

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Under most reasonable meanings of algebraic, no. Exponentials $+$ polynomials $=$ mess. –  André Nicolas Sep 28 '12 at 18:40
    
You might want take a look at Lambert's W. –  EuYu Sep 28 '12 at 18:52
    
if $x<0$ then $x+e^x<0+1$ and if $x>0$ we have $x+e^x>0+1$. This is an equation with unique solution, but it is hard to argue "algebraically" that $x+e^x$ is increasing thus 1 to 1... –  N. S. Sep 28 '12 at 19:55
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I'm teaching from Stewart this term and assigned this problem as homework. The point is that if you have an invertible function $f$ such that $f(0) = 1$ (such as the function $f(x) = e^x + x$), then you know that $f^{-1}(1) = 0$, even if you can't solve the equation $f(x) = a$ for $x$ algebraically in terms of $a$. –  Michael Joyce Sep 28 '12 at 19:56
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@MichaelJoyce And, we know it's invertible because it is increasing an increasing function. –  Graphth Sep 28 '12 at 21:52
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6 Answers

up vote 12 down vote accepted

Using the series expansion we have:

$$1+x+\frac{x^2}{2!}+ \dots + \frac{x^n}{n!}+\dots = 1-x$$

If $x$ is positive it is immediately obvious that there can be no equality.

If $x<0$ then the RHS is greater than 1 and $e^{x}<1$.

This is not strictly an "algebraic" solution, but with the term in $e^x$ we do not expect anything purely algebraic.

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0.o Good enough for my purposes. Thanks! –  YellPika Sep 28 '12 at 18:58
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Let $f(x) = e^x + x - 1$. Then, for any given $x$, $f(x) = 0$ if and only if $e^x + x = 1$.

You have already noticed that $f(0) = 1 + 0 - 1 = 0$, so it is a solution.

Now, we turn to calculus, not algebra. We have $f'(x) = e^x + 1$. Since $e^x > 0$ for all $x$, we know that $e^x + 1 > 0$ as well. In other words, $f'(x)$ is positive for all $x$ which tells us that $f(x)$ is an increasing function on the entire real line. Therefore, it could only possibly be 0 at one point, and you already found that point.

Now, if you haven't had calculus, you could still get the same basic idea. For example, you know $y = x$ is increasing. That is something you should know. Perhaps you have learned that $y = e^x$ is always increasing as well, because even in an algebra class, they would probably give you a bunch of properties of $y = e^x$ when they introduce it. Add these two functions together, and it's still increasing. Subtract 1, and the function is simply translated downward 1 unit, so it's still increasing everywhere. Again, the conclusion is the same.

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"Lambert W" is a hint for "algebraic solution".
The solution for $\mathrm{e}^x + x = 1$ is $1-\mathrm W(\mathrm{e})$,
to find ALL complex solutions, use all branches of the Lambert W ...

$$ \begin{align*} &\dots \\ 1 - \mathrm{W}_{-4}(\mathrm{e}) &= 3.159947300 + 23.47017395 i \\ 1 - \mathrm{W}_{-3}(\mathrm{e}) &= 2.849014724 + 17.17149358 i \\ 1 - \mathrm{W}_{-2}(\mathrm{e}) &= 2.393982241 + 10.86800606 i \\ 1 - \mathrm{W}_{-1}(\mathrm{e}) &= 1.532092122 + 4.597158013 i \\ 1 - \mathrm{W}_{0}(\mathrm{e}) &= 0.000000000 \\ 1 - \mathrm{W}_{1}(\mathrm{e}) &= 1.532092122 - 4.597158013 i \\ 1 - \mathrm{W}_{2}(\mathrm{e}) &= 2.393982241 - 10.86800606 i \\ 1 - \mathrm{W}_{3}(\mathrm{e}) &= 2.849014724 - 17.17149358 i \\ 1 - \mathrm{W}_{4}(\mathrm{e}) &= 3.159947300 - 23.47017395 i \\ 1 - \mathrm{W}_{5}(\mathrm{e}) &= 3.396557044 - 29.76478701 i \\ &\dots \end{align*} $$

explanation

$\mathrm{e}^x+x=1$
$\mathrm{e}^x=1-x$
$\mathrm{e} = (1-x)\mathrm{e}^{1-x}$
$\mathrm{W}(\mathrm{e}) = 1-x$
$x = 1-\mathrm W(\mathrm{e})$

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You can see this very easily graphically. The equation is $$e^x=1-x$$ and the two sides of the equation are plotted here (from Wolfram Alpha):enter image description here

The intuition for a formal proof also follows directly from the picture (the functions are both monotonic but in opposite directions), if that's your aim.

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Just to add weight to @Jonathan’s response: If $f$ and $g$ are an increasing and a decreasing function on $\mathbb{R}$ respectively, their graphs can cross at only one point. Inspection finds that point to be $(0,1)$, and you’re done.

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A semi-intuitive way to think about it: the right-hand side has no $e$, so whatever you plug into the $e^x$ has to kill the $e$. So it has to be $\ln a$ for some $a$. On the other hand, the right hand side has no $\ln$, so what you plug into the $x$ term must reduce to a non-$\ln$. If you try $\ln e^b$, you'll end up in a circle. If you think maybe things will cancel, you'll see that it wont work because you will always have a different number of "levels" of exponentials and logarithms. The only possibility is $\ln 1 = 0$, which quick inspection reveals that it works.

This is by no means a formal proof (at least as I've written it), but it provides the intuition to find the solution $x=0$ if you failed to notice it off the bat.

This also shows why the solution to something like $e^x + x = 2$ is not going to expressible in terms of functions like exponentials and natural logs.

And by the way, if you are also interested in complex solutions, all bets are off with this method, since $e^{x+2\pi i}=e^x$.

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