Which one is asymptotically larger?

What is the comparison of n and 2^logn?? I think both are asymptotically equal. I seen somewhere n > 2^logn if yes how?

2Comments
Vivek Vikram Singh vivek14 13 Jun 2016 05:37 pm

Yes both are equal. I hope you understand base of the log is 2. 

Lets assume

y = 2logn

log2y = log22logn

log2y = log2n . log22

log2y = log2n. 1 

log2y = log2n

So, y = n.

shivanisrivarshini shivanisrivarshini 13 Jun 2016 06:29 pm

n               2logn = 2log2n

n                n                  since  alogan =n

therefore both are same