$\phi(t):= \sqrt{2t |\log | \log t||} \; (t>0, t \neq 1)$ is equal to $t \mapsto \sqrt{2t \log \log 1/t}$ for small $t>0$

Question

I am looking at the function $$\phi(t):= \sqrt{2t |\log | \log t||} \; (t>0, t \neq 1).$$

How is this function equal to $t \mapsto \sqrt{2t \log \log t}$ for large $t$ and equal to $t \mapsto \sqrt{2t \log \log 1/t}$ for small $t$? I can see that if $t>e$, then $|\log |\log t|| = \log \log t$ so the first case holds. But I cannot see why for small $t>0$, we would have $|\log | \log t|| = \log \log 1/t$. I would greatly appreciate any help.

Answer 1

Note that $\log1/t=\log t^{-1}=-\log t$, so $\log\log1/t=\log(-\log t)$. Now if $t\leq 1$, then $\log t\leq 0$, so $|\log t|=-\log t$. Therefore $\log|\log t|=\log(-\log t)=\log\log1/t$.