# Asymptotic order

Below is a question I faced from an online test for preparation of exam and I had doubt in solution provided so I wanted to discuss my approach and ask about it. $$\frac{e^{n\log n}}n(A),n^{\sqrt n}(B),2^{n\log n}(C)$$ I want to order these functions based on increasing asymptotic order.

I rewrite A as $$\frac{n^{nlog e}}{n}=\frac{n^{1.44n}}{n}=n^{1.44n-1}$$

C is re-written as $$n^{nlog2}=n^n$$

So, finally, my order came to be A, B, C in increasing asymptotic growth.

Am I correct?

$$\frac{e^{n\log_2n}}n=\frac{e^{n(\ln n)/(\ln 2)}}n=\frac{n^{n/\ln 2}}n=n^{n/\ln 2-1}$$ $$2^{n\log_2 n}=n^n$$ We now compare powers: $$\sqrt n<\frac n{\ln 2}-1. Thus $$B, not $$A as you claimed.

• @Parcly-I think ln means log to the base e, but log is taken to be of base 2. – user3767495 Nov 17 '18 at 7:43
• @user3767495 My analysis would still hold in that case. – Parcly Taxel Nov 17 '18 at 7:44
• @Parcly-Thanks.I realised I was on the correct path, but somehow ordering in wrong way. – user3767495 Nov 17 '18 at 9:51