I have tries solving by taking log on both the sides, but was unsuccessful. Please upgrade my knowledge.
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2There are two ways of reading "10^10^10". Please clarify if it is (10^10)^10 or 10^(10^10) - or with formatting (useful link), $ (10^{10})^{10}$ or $ 10^{10^{10}}$ – Joffan Oct 25 '17 at 14:03
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Is $n$ an integer? – kingW3 Oct 25 '17 at 14:06
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3Note that the exponentiation is not associative, $a^{\left(b^c\right)}\neq \left(a^b\right)^c$. In any case $n^n = m$ is equivalent to $\exp\left(n\log n\right) = \exp\left(\log m\right)$ or to $n\log n=\log m$. The inverse function of $f(x)=xe^x$ is Lambert function $W(x)$, hence $f(\log n)=\log m$ leads to $n = \exp\left(W(\log m)\right)=\frac{\log m}{W(\log m)}$. – Jack D'Aurizio Oct 25 '17 at 14:21
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$$n=1,105,747,502.593426$$ – Raffaele Oct 25 '17 at 16:10
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What's the calculus tag got to do with this? – Simply Beautiful Art Oct 26 '17 at 01:28
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@Joffan: Its 10^10^10 without any bracket, Later one in your comment. – BlackBrain Oct 26 '17 at 03:22
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We have $$n^n=10^{(10^{10})}\\ n \log_{10}n=10^{10}$$ Now without the Lambert W function you won't get an exact solution. Alpha gives $n = e^{W(10000000000 \log(10))} \approx1.1$ billion. You can find the numeric value by iteration. We can guess that $n$ is about $1$ billion, then iterate $n=\frac {10^{10}}{\log_{10} n}$ to convergence. We choose that form because $\log_{10}n$ is slowly varying, so the convergence will be rapid.
Ross Millikan
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