# Solve $x^x = a$ for known $a$? [duplicate]

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For example if you have $x^x = 2$, can you express $x$ as a numerical expression containing only the addition, multiplication and exponentiation operators?

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## marked as duplicate by J. M., Did, Jim, Stefan Hansen, Lord_FarinJun 12 '13 at 8:11

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Not really – Ilya Apr 4 '13 at 12:19
See here‌​. – Mhenni Benghorbal Apr 4 '13 at 12:21
The solution of your equation is $\frac{\ln(a)}{W(\ln(a))},$ where $W$ is the Lambert $W$-function. – Mhenni Benghorbal Apr 4 '13 at 12:42

## 1 Answer

I looked at problems like this once and found the Lambert W function. It is not the addition, multiplication and exponential operators you have asked for. I can't see how you would do that since we get the exact answer as follows:

$x\exp(x)=y \iff x=W(y)$

$x^x=2$

$x \log x=\log 2$

$(\log x)\exp(\log x) = \log 2$

$x=\exp(W(\log 2))$

But $\log(2)=W(\log 2)\exp(W(\log 2))=W(\log 2)x$

$x=\frac{\log 2}{W(\log 2)}$

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I don't really know this function but is there a method to evaluate this at a point? I mean if it has a simple taylor expansion then I guess we can express it using basic operations. – Alraxite Apr 4 '13 at 17:47
From the Wikipedia page $W_0 (x) = \sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}\ x^n = x - x^2 + \frac{3}{2}x^3 - \frac{8}{3}x^4 + \frac{125}{24}x^5 - \cdots$ – shilov Apr 5 '13 at 0:41