Because $\log _{ 10 }{ 100 } =2$

Calculus Level 3

Compare $\displaystyle A= \sum _{x=1}^a \left(\log _{\sqrt a}(x) -1\right)^2$ and $B=\dfrac a2$ .

P/S: Thanks for the reports. I had this idea when I was reading one of them.

A=B

A<B

It depends on

a

A>B

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