asymptotic notations

Question

√logx  = O(loglogx) is it true or false?

and explain why?

Comments

False. for x>0 =>$\sqrt{\log x}$ > $\log\left(\log x\right)$

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take x = 2¹⁰²⁴

√logx = √log₂2¹⁰²⁴   = 2⁵¹²  

log₂log₂x  = log₂log₂ 2¹⁰²⁴

                 =log₂1024

                 =10

So we can say that √logx > log₂log₂x  and hence ur given statement is false

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√logx= O(loglogx)

apply log on both sides

1/2* logx <= logloglogx

which is false

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False.

 √logx > log₂log₂x