# Asymptotic growth of function

## Question

I want to arrange which is asymptotically faster -:

$$n^{100},2^{n},n^{\log\,n}$$

## My approach

I know that Exponential function will beat Polynomial function from here

but i am thinking of different way of solving it.

let $$f_{1}=n^{100},f_{2}=2^{n},f_{3}=n^{\log n}$$

Take log both sides-:

$$y_{1}=\log f_{1}=100 *\log\,n$$ $$y_{2}=\log f_{2}=n$$

$$y_{3}=\log \,f_{3}=\log\,(\log\,n*n)$$

hence $$f_{2}>f_{3}>f_{1}$$

Am i right?

• Yes, this is correct. It works because of the continuity of $\log$ and $\exp$. $$\log f\leq\log g\iff \log\frac fg\leq 0\iff f\leq g$$ Oct 6, 2017 at 19:29
• It seems more straight forward to rewrite your functions using $x = e^{\ln x}$. Then you're just comparing $\{e^{100\ln n}, e^{(\ln 2) n}, e^{(\ln n)^2}\}$. Oct 6, 2017 at 19:31
• @PrasunBiswas No, it's not correct.
– zhw.
Oct 6, 2017 at 19:42

Take logarithms of the three quantities, it is then clear that (for sufficinetly large $n$) \begin{eqnarray*} 100 \ln n < (\ln n)^2 < n \ln 2. \end{eqnarray*} So your answer should be $f_1 < f_3 < f_2$.
• That answer has $f_1> f_3,$
• @DonaldSplutterwit can you please explain that after taking log of $f_{3}$, it is $(\ln n )^{2}$.I am okk upto $\log \log n \, n$ Oct 6, 2017 at 19:46
• $\ln (n^m) = m \ln n$ ... now let $m = \ln n$ ... so its $\ln n \times \ln n$. Oct 6, 2017 at 19:49