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Re-write $(\log(n))^{\log n}$ as $n^{\log(\log n)}$

I have not managed to re-write it, the only rule I thought of is to write it as $\log(n)\log(n)$ but I've no idea how to bring the $n$ as the basis.

Adi Dani
  • 16,949

4 Answers4

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Note that, by definition, $a^b=e^{b\ln a}$, so you get $$\left(\ln x\right)^{\ln x}=e^{\ln x \ln\ln x}=\left(e^{\ln x}\right)^{\ln\ln x}=x^{\ln\ln x}.$$

heptagon
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$$\log(n)^{\log(n)}=e^{\log(\log(n)^{\log(n)})}$$ $$e^{\log(\log(n)^{\log(n)})}=e^{\log(n)\log(\log(n))}$$ $$e^{\log(n)\log(\log(n))}=n^{\log(\log(n))}$$

user2825632
  • 2,881
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Take log of both expressions.

For the first expression you get $\log((\log n)^{\log n})=\log n\log\log n$ since $\log(a^b)=b\log a$.

For the second expression you get $\log(n^{\log\log n})=\log\log n\log n$.

So the two expressions are equal.

pi66
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In general (for $a > 0$, if you don't want to get into complex numbers) $$a^c = \exp(c \log a)$$ So (using that twice, once with $a = \log n$ and once with $a = n$) $$ \eqalign{(\log n)^{\log n} &= \exp((\log n)(\log(\log n)))\cr &= \exp((\log(\log n))(\log n))\cr &= n^{\log(\log n)}}$$

Robert Israel
  • 448,999