Time complexity in algorithm

Question

(logn)^k<=cn=logn<=c(n)^ 1/k how??

Answer 1

Lets say n=2^m Also assume log as log₂

A=(lgn)^k=(lg(2^m))^k=m^k

B=cn=c*2^m

C=lgn=lg(2^m)=m

D= c(n) ^1/k=2^m/k

 

C<=A<=D<=B

Comments

Thanks @saksham for giving explanation ,I got it..

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I think if we take log both sides then it would be clear which one is greater ..

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Yes, actually the questions should specify what k is
Because for smaller value of k the result I showed need not be true

Yes Log works an done more thing u can try is google any online function plotting website and plot functions.