# How to Solve $x^2=2^x$ [duplicate]

I was thinking of random math problems to myself when I came across this:

$x^2 = 2^x$

It seemed very simple at first, but after trying multiple ways to solve it, I had no idea what to do.

So, I began doing a lot of searches for this problem, and I found out that in order to solve $x^2 = 2^x$, you have to use some function of which I have never heard before. So, I have three main questions regarding this problem.

1) is it possible to solve $x^2 = 2^x$ arithmetically? If so, how would one do so? If not, why not? In other words, why would it be impossible to solve this problem arithmetically?

2) Why is it that addition and multiplication are commutative, but exponentiation is not? I understand not all forms of multiplication in mathematics are non-commutative, but for the sake of this problem, exponentiation feels like the "next step" right after multiplication. Here's what I mean by this:

Multiplication is just adding a number $N$ times to a number.

Exponentiation is just multiplying a number $N$ times by a number.

3) Would it be possible to invent your own function to solve $x^2 = 2^x$, or rather, $x^y = y^x$? And after you found a way to represent your answer, how could you go about evaluating your answer without using a calculator?

If anything doesn't make sense above, I can do my best to clarify. I also know some of these questions have been asked before, but I didn't really understand any of the answers, and the questions weren't as specific as mine.

## marked as duplicate by Joel Reyes Noche, tatan, Parcly Taxel, user99914, user91500Sep 30 '16 at 9:06

• Take logarithm both side it may help to solved this. – Sufaid Saleel Sep 30 '16 at 6:16
• Has this question been asked before? I do remember seeing it but I can't find it... – Parcly Taxel Sep 30 '16 at 6:19
• Lambert W function...for the general case, but $2$ and $4$ are two obvious solutions. The third solution is not so obvious – ClassicStyle Sep 30 '16 at 6:21
• – dxiv Sep 30 '16 at 6:25
• @dxiv: The second question isn't covered by those posts. – celtschk Sep 30 '16 at 6:42

Sooner or later, you will learn that the solution(s) of $$x^y=y^x$$ express in terms of Lambert function and the solution write $$x=-\frac{y}{\log (y)} W\left(-\frac{\log (y)}{y}\right)$$ It cannot be expressed in terms of elementary function.

In the real domain, function $W(z)$ is only defined for $z \geq -\frac 1e$ and, as you will see in the Wikipedia page, for $-\frac 1e \leq z \lt 0$, it i multivalued (upper and lower branches).

For computing the values, the Wikipedia page contains expansions which are very good.

Out of curiosity, search on this site and you will find many uses and applications of the beautiful Lambert function.

You can find a couple integer solutions, $x = 2$, $x = 4$, by just guessing.

There is another solution $x \approx -0.766665$, but this requires the Lambert W function.

$$x = -\frac{2W\bigl(\frac{\ln 2}{2}\bigr)}{\ln 2}$$

For a similar question, see here: Solve for $x$: $2^x = x^3$

It's $\displaystyle x_1^{x_2}=x_2^{x_1}$ with $\displaystyle x_1=(1+\frac{1}{t})^t$, $\displaystyle x_2=(1+\frac{1}{t})^{t+1}$ and therefore $1<x_1<e<x_2$.

With $t=1$ we get $2^4=4^2$.

$x^2=2^x$ with $x<0$ means $y^\frac{1}{y}=\frac{1}{\sqrt{2}}$ with $y:=-x$.

You can use the Lambert W-function or simply the iteration $y_{n+1}=(\frac{1}{\sqrt{2}})^{y_n}$ with $y_0:=1$.