miniliv4

2022-09-06

Why does ${n}^{\mathrm{ln}\mathrm{ln}n}=(\mathrm{ln}n{)}^{\mathrm{ln}n}$?

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:

$${n}^{\mathrm{ln}\mathrm{ln}n}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}(\mathrm{ln}n{)}^{\mathrm{ln}n}$$

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.

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:

$${n}^{\mathrm{ln}\mathrm{ln}n}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}(\mathrm{ln}n{)}^{\mathrm{ln}n}$$

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

Beginner2022-09-07Added 8 answers

Try put the two expressions into the form ${e}^{(\cdots )}$ and compare.

Formula : ${x}^{y}={e}^{y\mathrm{ln}(x)}$

Formula : ${x}^{y}={e}^{y\mathrm{ln}(x)}$

Bruce Sherman

Beginner2022-09-08Added 2 answers

Take the logarithm of both expressions.

$\mathrm{ln}({x}^{y})=y\cdot \mathrm{ln}(x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}(\mathrm{ln}(n))\cdot \mathrm{ln}(n)=\mathrm{ln}(n)\cdot \mathrm{ln}(\mathrm{ln}(n))$

$\mathrm{ln}({x}^{y})=y\cdot \mathrm{ln}(x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}(\mathrm{ln}(n))\cdot \mathrm{ln}(n)=\mathrm{ln}(n)\cdot \mathrm{ln}(\mathrm{ln}(n))$

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