1. n > (log n)log n
take log on both side
we get logarithmic value of n is log n.............i
logarithmic value of (log n)log n is log n(log log n)..................ii
Now put n=22^x
putting it in i we get log n =2x
putting it in ii we get (log n)log n = log n(log log n) =2x . x
then n < (log n)log n
2. take log in both the function
2n taking log value will be n log 2 =n .................i
n√n taking log value will be √n log n ...................ii
again take n=2x
2n =n log 2 =n taking n=2x value will be 2x ......................iii
n√n = √n log n taking n=2x value will be 2x/2 . x ............iv
Here 2x > 2x/2 . x
So, 2n > n√n
3. take log on both side
2n taking log value will be n
nlog n taking log value will be log n . log n =(log n)2
Here too linear value of n will be greater than logarithmic value of (log n)2
So, 2n > nlog n