Comparing functions that have logs in exponents

Question

I'm not that familiar with logs in general so not sure how to handle when say comparing two functions to see which one would grow slower / faster

$$n^{\log\log n}$$

to this...

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

Anyone able to help clarify? Just not sure what I should be doing when a log is in the exponent. I've only dealt with functions that have a base that is the same as the base of the function. For example...

$$2^{\log_2 9} = 9$$

Answer 1

You want to compare which of the functions $f(n)=n^{\log \log n}$ and $g(n)=(\log n)^{\log n}$ grows faster. To determine this, let's look at their ratio (where $\log$ is assumed to be the natural logarithm and $t=\log n$) $$\frac{f(n)}{g(n)}=\frac{n^{\log \log n}}{(\log n)^{\log n}}=\frac{(e^t)^{\log t}}{t^t}=\frac{e^{t \log t}}{e^{t \log t}}=1,$$

and hence they grow at the same rate!

To show $t^t=e^{t \log t}$, let's solve $$t^t=e^{kt}=(e^k)^t \implies t=e^k \implies k=\log t.$$

Answer 2

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

so they are the same thing.