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I am having difficulty understanding how this follows.

$(\log n)^{ (\log n) } = 2^{(\log n)(\log (\log n))} = n^{\log \log n}$

Which logarithmic identities are used to go through each equality?

e.g. how do you first go from

$(\log n)^{ (\log n) } = 2^{(\log n)(\log (\log n))}$

and then to

$2^{(\log n)(\log (\log n))} = n^{\log \log n}$

(The log base must be 2 or else this equality won't hold)

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    @DonAntonio For clarification, the edits occurred after I asked if the base is 2.2012-07-28

3 Answers 3

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I'm assuming the base is $2.$ Otherwise the equality doesn't hold.

$2^{(\log n)(\log (\log n))} = 2^{(\log (\log n)) (\log n)} = (2^{(\log (\log n))})^{(\log n)}$ and $ 2^{(\log (\log n))} = \log n $ because $ 2^{\log x} = x.$


Edit: to reflect the update in the question:

$ 2^{(\log n)(\log (\log n))} = (\color{blue}{2^{(\log n)}})^{(\log (\log n))} = \color{blue}{n}^{{(\log (\log n))}} = n^{\log \log n} $ because again $ 2^{\log n} = n.$

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    @J.D. , isn't your answer taking $\,\log =\log_2\,$ ?2012-07-28
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For any $1\neq a\in\Bbb R^+\,\,,\,\,x^y=a^{y\log_ax}\,\,,\,\text{whenever LHS is defined. }$

Thus, $(\log n)^{\log n}=2^{\log n\,\log_2(\log n)}$

So your equality follows if the logarithm here is taken in base $\,2\,$ and not $\,10\,$ , as you wrote...

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Let

$y=\log_2^{\log_2 n} n$

Taking the logarithm (in base 2) of both sides

$\log_2 y=\log_2 n \log_2 \log_2 n$

Now, remember that $2^{\log_2 n}=n$. Thus

$y=2^{\log_2 y}=2^{\log_2 n \log_2 \log_2}$

Also recall that $a^{bc}=(a^b)^c$. Thus

$y=2^{\log_2 n \log_2 \log_2 n}=(2^{\log_2 n})^{\log_2 \log_2 n}=n^{\log_2 \log_2 n}$