Let $L(X) = \exp(\sqrt{\log X \log \log X})$
Prove that if $c > 0$,$ Y = L(X)^c$, and $u = \log X/ \log Y$ , then $$u^u = L(X)^{(1/2c)(1+o(1))}$$
I've tried to write $u^u = (\log X/ \log Y)^{\log X/ \log Y}$
But that doesn't seem to get me anywhere. Any help is appreciated!
First, we have $\ln u^u = u \ln u = \frac{1}{\ln Y} \left(\ln X \ln \frac{\ln X}{\ln Y} \right) = \frac{1}{\ln Y} \left(\ln X (\ln \ln X- \ln \ln Y) \right)$, and $\ln \ln Y = \ln (c\cdot \sqrt{\ln X \ln \ln X}) =\frac{1}{2} \ln \ln X + o(\ln \ln X), \ as \ X \rightarrow +\infty.$ Therefore, $\ln u^u = \frac{1}{2c \sqrt{\ln X \ln \ln X}} \ln X \ln \ln X \cdot (1+ o(1) ).$ On the other hand, we have $\ln L(X) = \sqrt{\ln X \ln \ln X}$, and so $$ \frac{\ln u^u}{ \ln L(x)} =\frac{1}{2c} (1 + o(1)). $$