Is there any "superlogarithm" or something to solve an equation like this:
$$x^x = 10?$$
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1$\begingroup$ Newton's method works for (almost) every equations and is pretty fast (and works better for convex functions like yours). $\endgroup$– user10676Jun 11, 2013 at 17:21
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21$\begingroup$ It's usually called the Lambert W, but superlogarithm is a far better name. $\endgroup$– LucasJun 11, 2013 at 21:23
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1$\begingroup$ In Mathematica - and no where else, as far as I know - the Lambert W is referred to as the Product Log. $\endgroup$– horchlerJun 12, 2013 at 0:02
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$\begingroup$ A related problem. $\endgroup$– Mhenni BenghorbalJun 12, 2013 at 0:10
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3 Answers
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What you're looking for is the Lambert $W$ function. This is the function such that: $$W(xe^x) = x$$
It does not have a "simple" or explicit form.
To solve your equation, we follow this process: $$x^x = 10$$ $$x\ln x = \ln10$$ $$e^{\ln x}\ln x = \ln10$$ $$W(e^{\ln x}\ln x) = W(\ln10)$$ $$\ln x = W(\ln10)$$ $$x = e^{W(\ln10)}$$
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Yes - it's called the Lambert W Function. Scroll down and take a look at Example 2.
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The Lambert $W$ function is technically a multi-valued function. Equations such as yours can also be solved with a simpler single valued function, the Wright $\omega$ function (see also Corless & Jeffrey 2002), which is defined with respect to the Lambert $W$. The Lambert $W$ by it's definition is the natural choice for exponential functions like yours, while the Wright $\omega$ is better suited to logarithmic ones (e.g., $z = \omega+\text{ln}(\omega)$), but one can often transform between the two. In your case, the two are equivalent via:
$$x^x = a \rightarrow x = \frac{\ln a}{W_0(\ln a)} = \frac{\ln a}{\omega(\ln(\ln a))}$$
for $a \in \mathbb{Z}$ and where $W_0$ is the upper (principal) branch of the Lambert $W$ function.
If you happen to need numeric values, the Wright $\omega$ function may be easier and more efficient to calculate. For $z = \ln(\ln 10)) \approx 0.834032...$ (and real $z \in [-1, 2]$), the following is an approximation for $\omega(z)$:
$$\omega(z) \approx 1 + \frac{1}{2}(z−1) + \frac{1}{16}(z−1)^2 − \frac{1}{192}(z−1)^3 − \frac{1}{3072}(z−1)^4 + \frac{13}{61440}(z−1)^5...$$
Matlab's Symbolic Toolbox has builtin support via wrightOmega, as does Maple with Wrightomega. Additionally, I've implemented the Wright $\omega$ function for complex double-precision evaluation in Matlab based on a 2012 paper by Lawrence, Corless, and Jeffrey – it's 3 to 4 orders of magnitude faster than performing variable precision/symbolic calculations.
