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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?

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

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

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