Identity with logarithms? Is it correct?
$$(\log\,n)^{(\log\,n)} = n^ {(\log\,\log\,n)} $$
If yes and they are equal, how can I get $(\log n)^{\log n}$ from $n^{\log \log n}$ ?
Thanks.
 A: Yes. We have
$$(\log n)^{\log n}=\exp(\log n\log(\log n))=\exp(\log n^{\log(\log n)})= n^{\log(\log n)}$$
A: Yes. I'll assume that you are using logarithms with a default base of $2$. Observe that:
\begin{align*}
n^{\log \log n}
&= 2^{\log (n^{\log \log n})} &\text{since logs and exponentials are inverses of each other} \\
&= 2^{(\log \log n)\log n} &\text{using the power rule for logs} \\
&= 2^{(\log n)\log (\log n)} &\text{by the commutativity of multiplication} \\
&= 2^{\log ((\log n)^{\log n})} &\text{using the power rule for logs} \\
&= (\log n)^{\log n} &\text{since logs and exponentials are inverses of each other}
\end{align*}
A: Stop thinking and write both definitions:
$$
LHS = \log n^{\log n} = \exp(\log n \times \log(\log n))
\\
RHS = n^{\log\log n} = \exp(\log \log n \times \log n)
\\
\implies RHS = LHS
$$
A: A symmetrical approach:
$$\begin{align}
(\log a )(\log b)&=(\log b )(\log a)\\
\log (b^{\log a})&=\log (a^{\log b})\\
b^{\log a}&=a^{\log b}\\
\end{align}$$
Put $b=\log n$ and $a=n$:
$$(\log n)^{\log n}=n^{\log(\log n)}$$
Valid for any base.
A: Use these two properties:
$$\large a^b=c^{b(\log_{\; c}a)}\tag{1}$$
$$\large \log_ab=\frac{\log_c b}{\log_c a}\tag{2}$$
You can write:
$$\require{cancel}\Large (\log\,n)^{(\log\,n)} \\\Large =^{(1)} n^{(\log\;n)\log_n(\log\; n)}\\\Large=^{(2)} n^{\cancel{(\log\;n)}\frac{(\log\log\; n)}{\cancel{\log\;n}}}\\\Large=n^{\log\;\log\;n}$$  
