miniliv4

2022-09-06

Why does ${n}^{\mathrm{ln}\mathrm{ln}n}=\left(\mathrm{ln}n{\right)}^{\mathrm{ln}n}$?
Note: this is (part of my solution to) a homework question. Please DO NOT tell me the answer!
I am trying to compare the following functions:
${n}^{\mathrm{ln}\mathrm{ln}n}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left(\mathrm{ln}n{\right)}^{\mathrm{ln}n}$
It appears that they are equal (assuming $n>1$), but I have absolutely no idea why this would be the case.
I am missing something really obvious and I have been hitting my head on this for about 45 minutes. A hint would be appreciated.

Lamar Esparza

Try put the two expressions into the form ${e}^{\left(\cdots \right)}$ and compare.
Formula : ${x}^{y}={e}^{y\mathrm{ln}\left(x\right)}$

Bruce Sherman

Take the logarithm of both expressions.
$\mathrm{ln}\left({x}^{y}\right)=y\cdot \mathrm{ln}\left(x\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}\left(\mathrm{ln}\left(n\right)\right)\cdot \mathrm{ln}\left(n\right)=\mathrm{ln}\left(n\right)\cdot \mathrm{ln}\left(\mathrm{ln}\left(n\right)\right)$

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