Hi all ---

For this multiple choice question, I don't know how to compare (a) and (b).

I try --- $\displaystyle \log \pi ? \sqrt{\log (\pi)^2}$

Then --- $\displaystyle \log \pi ? \sqrt{2\log\pi}$

or if I sub in L --- $\displaystyle L ? \sqrt{2L}$

Squaring both sides --- $\displaystyle L^2 ? 2L$

Because $\displaystyle 0 < L < 1 $ --- squaring this for the 1st and then multiplying by 2 for the 2nd --- $\displaystyle 0 < L^2 < 1$ and $\displaystyle 0 < 2L < 2$.

1st Question ---But how can I compare the last two inequalities? I don't know what specifically L is?

2nd Question ---How does solution get $\displaystyle \sqrt{2L} > \sqrt{L \times L}$?

Thanks a lot ---