Logarithm Raised to a Power of a Logarithm $(\log n)^{\log n}$. I am trying to understand how $(\log n)^{\log n} $ is equivalent to $ n^{\log\log n} $ with respect to how it is derived and the associated logarithm properties/rules that make this possible.
There is a solution located here which appears to be applicable.
The solution provided at the above location is:
$$\log(n)^{\log(n)} = \left ( b^{\log(\log(n))} \right )^{\log(n)}=(b^{\log(n)})^{\log(\log(n))}=n^{\log(\log(n))}$$

assuming the log is base $b$. Since $\log(\log(n))$ is growing, this grows faster than just $n$ which has a fixed exponent.


*

*Is this the correct derivation/solution?

*If so, would one be able to provide more details along with the associated logarithmic properties/rules on how this is derived? I have looked in my pre-calculus book from a few years ago to find the applicable logarithm properties/rules but cannot find them or a similar example.

*How does $(\log n)^{\log(n)} = ( b^{\log(\log(n))} )$?

*How does $ ( b^{\log(\log(n))} ) = (b^{\log(n)})^{\log(\log(n))} $?

*How does $ (b^{\log(n)})^{\log(\log(n))} = n^{\log(\log(n))}$?

Any help on this would be greatly appreciated.
Thank you very much.
 A: It appears you're assuming the base of the logarithms is $b,$ but leaving this out of your expressions makes things confusing to follow, since it's essentially universal that $\log$ means either base-$e$ logarithm or base-$10$ logarithm. Initially I thought there was a typo when $b$ suddenly showed up out of the blue in an expression.
I also find that the type of derivation you have difficulty with being something that is not very "discoverable" in the form shown, since the manipulations are not those one ordinarily makes use of in practice when working with logarithms (unless maybe doing this specific type of problem). For me, it's much more natural to introduce an equation with one side being the expression and, for the other side, introduce some symbol the expression is set equal to, after which you variously rewrite the equation using more standard (and thus more familiar) logarithm properties.
Begin by writing $\;u = (\log_b n)^{\log_b n}.\;$ Now take logarithm-to-base-$b$ of both sides and use basic (non-tricky) logarithm properties:
$$  \log_b u \;\; = \;\; \log_b \, \left[ (\log_b n)^{\log_b n} \right] $$
$$  \log_b u \;\; = \;\; (\log_b n) \cdot \log_b \, [\log_b n] $$
Now solve for $u$ by using the property/definition that $\;``\log_b u = \text{stuff}"\;$ is equivalent to $\;``u = b^{\text{stuff}}"$ (here, for instance) to obtain
$$  u \;\; = \;\; b^{(\log_b n) \cdot \log_b \, [\log_b n]} $$
Recall that $\;b^{(\log_b n)} = n,\;$ this being one of the two fundamental identities for the fact that exponentiation-base-$b$ and logarithm-base-$b$ are inverse functions, namely $\;b^{(\log_b x)} = x\;$ and $\;\log_b(b^x) = x.\;$ The most recent displayed equation above appears to have $\;b^{(\log_b n)}\;$ in it, but not as an isolated term that we can replace with $n.$ However, by using the property $\;A^{BC} = (A^B)^C\;$ we can rewrite that equation in such a way that $\;b^{(\log_b n)}\;$ shows up in a way that allows for replacement by $n.$
$$  u \;\; = \;\; b^{(\log_b n) \cdot \log_b \, [\log_b n]} \;\; = \;\; {\left( b^{(\log_b n)}\right)}^{\log_b \, [\log_b n]}  $$
$$  u \;\; = \;\; n^{\log_b \, [\log_b n]} $$
