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(logn)^k<=cn=logn<=c(n)^ 1/k how??
in Algorithms retagged by
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Lets say n=2m Also assume log as log2

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

B=cn=c*2m

C=lgn=lg(2m)=m

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

 

C<=A<=D<=B

4 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.
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