why $n^{\log{\log{n}}}=\log{n}^{\log{n}}$?

Question

I was reading the solution of one exercise on my book on algorithms. What I need to do in short is order some function from fast to slow. This is the link of the solution: site.

At some point I get the following property: $$n^{\log{\log{n}}}=\log{n}^{\log{n}}$$ but how do you prove that?

I've tried to use the properties of logarithm, but I wasn't able to get that result. Can you give me some hints?

Answer 1

$$n^{\log{\log{n}}} = n^{(\log{\log{n}})/(\log{n})(\log n)} = n^{(\log_n \log n)(\log{n})} = \left(n^{\log_n \log n}\right)^{\log{n}} = (\log n)^{\log{n}}$$

Answer 2

Let $$x = \log n^{\log n}.$$ Then, $$ e^{\log x} = e^{\log\log n^{\log n}} = e^{\log n\log\log n} = n^{\log\log n}.$$

Answer 3

$$n^{\ln x}=x^x, x=\ln n$$

$$\implies \ln n^{\ln x}=\ln x^x$$

$$\implies \ln x \ln n=x \ln x$$

$$\implies x=\ln n$$

Reverse the steps and generate the formula.