time complexity

Question

1. n > (log n)log n

2. 2n > n√n

3. 2n > nlog n

explain plz

Answer 1

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=2^{2^x}

putting it in i we get log n =2^x

putting it in ii we get (log n)^{log n}  = log n(log log n) =2^x . 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=2^x

2ⁿ =n log 2 =n taking n=2^x value will be 2^{x ......................}iii

n^√n = √n log n taking n=2^x value will be 2^x/2 . x ............iv

Here 2^x > 2^x/2 . x 

So, 2ⁿ > n^√n

3. take log on both side

2ⁿ   taking log value will be n

n^{log 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 > n^{log n}