Identity with logarithms?

Question

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.

Answer 1

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: $$\require{cancel}\Large (\log\,n)^{(\log\,n)} \\\Large =^{(1)} n^{(\log\;n)\log_n(\log\; n)}\\\Large=^{(2)} n^{\cancel{(\log\;n)}\frac{(\log\log\; n)}{\cancel{\log\;n}}}\\\Large=n^{\log\;\log\;n}$$

Answer 2

Yes. We have

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

Answer 3

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*}

Answer 4

Stop thinking and write both definitions:

$$ LHS = \log n^{\log n} = \exp(\log n \times \log(\log n)) \\ RHS = n^{\log\log n} = \exp(\log \log n \times \log n) \\ \implies RHS = LHS $$

Answer 5

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.