Comparing $x\log x^2$ and $(\log x)^{2\log x}$.

Question

I have this question in a past worksheet that asks me to order 2 functions by order of growth:
$$f_{1}(x) = x\log(x^{2})\\ f_{2}(x) = (\log (x))^{2\log(x))} $$ I had seen in desmos the $f_{2}(x)$ is greater, but I'm not sure how to prove this, could I do it with little o? I'm also thinking that as the second function has a power containing $x$ it can be thought of as exponential.
Thanks

Answer 1

We have $f_1(x)=2x\log x$, while $f_2(x)=e^{2\log x\log \log x}$ by the fact that $a^x = e^{x\log a}$.

So $f_2(x) = (x^2)^{\log \log x}$, which is clearly larger due to the squared term.