# Thread: Which one is bigger between 2^n and n^logn ?

1. ## Which one is bigger between 2^n and n^logn ?

Hi

Which one is bigger between 2^n and n^logn ?
How can I know?
But, still cannot get an answer..

2. ## Re: Which one is bigger between 2^n and n^logn ?

Since $2^x = e^{x\log(2)}$ and $x^{\log x}= e^{(\log x)^2}$, then $2^x \ge x^{\log x}$ when $x\log 2 \ge (\log x)^2$.

Let $f(x) = x\log(2) - (\log x)^2$. You can easily calculate $f(1) = \log(2) > 0$. Then, $f'(x) = \log(2) - 2\dfrac{\log x}{x}$. $f'(1) = \log(2)$. You want to show that $f'(x) \ge 0$ for all $x>1$. Suppose $f'(x) = 0$. Then $\log(2) = 2\dfrac{\log x}{x}$. This happens when $x\log 2 = 2\log x$. Since $2^x > x^2$ for all $x>1$, $f'(x)>0$ for all $x>1$. So, $2^x>x^{\log x}$ for all $x\ge 1$.

3. ## Re: Which one is bigger between 2^n and n^logn ?

Oh, I wasn't thinking. You are correct that $x^2>2^x$ for $2. So, $f(x)$ has local extrema at $x=2,x=4$. Then $f(2) = 2\log(2) - (\log 2)^2>0$ and $f(4) = 4\log(2) - (\log 4)^2>0$, so since $f(x)$ increases everywhere else, it must be greater than zero for all $x>1$.