I'm new to proofs and although I find this proof fascinating, I'm hoping someone could help me understand how the seasoned mathematician arrived at this result... I'm assuming it involves some abstraction of the Lambert W function.

This is a solution from the Book of Proof by Professor Richard Hammack.

Prove the equation $x^2 = 2^x$ has three real solutions


By inspection, the numbers $x=2$ and $x=4$ are two solutions of this equation. But there is a third solution. Let $m$ be the real number for which $m2^m = \frac{1}{2}$. Then the negative number $x=-2m$ is a solution, as follows.

$x^2=(-2m)^2 = 4m^2 = 4(\frac{m2^m}{2^m})^2 = 4(\frac{\frac {1}{2}}{2^m})^2 = 2^2\cdot[ 2^{-(m+1)}]^2 = 2^x$.

Therefore we have three solutions $2$, $4$ and $m$. $\Box$

Again, I understand the calculations within the proof. But what mental process might unfold to make this realization?

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    $\begingroup$ You are not asking about this, but I find this proof incomplete. I think that the sentence "Let $m$ be the real number for which $m2^m=1/2$" should be justified. $\endgroup$
    – ajotatxe
    Sep 7, 2016 at 20:06
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    $\begingroup$ The conclusion on this proof should be "Therefore we have atleast three solutions" as no arguments are presented that rule out the existence of other solutions. $\endgroup$
    – Winther
    Sep 7, 2016 at 20:22
  • $\begingroup$ @Winther: I was wondering about this myself as I read through it. $\endgroup$ Sep 7, 2016 at 20:26
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    $\begingroup$ @DaveL.Renfro On second though I think it is acctually correct as written since the problem just asks to show that we have three solutions and the solution does just that. $\endgroup$
    – Winther
    Sep 7, 2016 at 20:32
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    $\begingroup$ @Winther: Yes, your second thought does fix things. For what it's worth, this is why you see "a total of" or "exactly" appearing often in math problems in high quality high-stakes tests. $\endgroup$ Sep 7, 2016 at 21:31

2 Answers 2


This is a terrible argument. The desired solution pops out of nowhere in a totally unnecessary way. A much more conceptual approach, which you could take without knowing in advance what answer it will spit out, is to think about what the graphs of $y = x^2$ and $y = 2^x$ look like (because you're trying to figure out when they cross). The first graph is a parabola, and the second graph is an exponential. At $x = 0$, the parabola is below the exponential, and at $x = 2$, they meet. Since exponentials eventually grow much faster than polynomials, there must be at least one more positive solution when the exponential overtakes the parabola again, which turns out to be $x = 4$.

But there must also be at least one more negative solution, where the parabola crosses the exponential while it's exponentially decaying. It doesn't particularly matter what the exact form of this solution is; you don't really learn anything from expressing it in terms of the Lambert W function. It's straightforward to narrow down an interval it must belong to using the intermediate value theorem; for example, it must be between $-1$ and $0$, and even must be between $-1$ and $- \frac{1}{2}$.

You can actually see the graph at WolframAlpha, which instantly tells you with no work that there are three solutions. With a little more work you can even prove that the only three solutions you can see on this graph are in fact the only solutions.


I like this video which explains how to drive all solutions:

ALL solutions to x^2=2^x

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