Hi
Which one is bigger between 2^n and n^logn ?
How can I know?
I try to put a log in both sides.
But, still cannot get an answer..
Since $\displaystyle 2^x = e^{x\log(2)}$ and $\displaystyle x^{\log x}= e^{(\log x)^2}$, then $\displaystyle 2^x \ge x^{\log x}$ when $\displaystyle x\log 2 \ge (\log x)^2$.
Let $\displaystyle f(x) = x\log(2) - (\log x)^2$. You can easily calculate $\displaystyle f(1) = \log(2) > 0$. Then, $\displaystyle f'(x) = \log(2) - 2\dfrac{\log x}{x}$. $\displaystyle f'(1) = \log(2)$. You want to show that $\displaystyle f'(x) \ge 0$ for all $\displaystyle x>1$. Suppose $\displaystyle f'(x) = 0$. Then $\displaystyle \log(2) = 2\dfrac{\log x}{x}$. This happens when $\displaystyle x\log 2 = 2\log x$. Since $\displaystyle 2^x > x^2$ for all $\displaystyle x>1$, $\displaystyle f'(x)>0$ for all $\displaystyle x>1$. So, $\displaystyle 2^x>x^{\log x}$ for all $\displaystyle x\ge 1$.
Oh, I wasn't thinking. You are correct that $\displaystyle x^2>2^x$ for $\displaystyle 2<x<4$. So, $\displaystyle f(x)$ has local extrema at $\displaystyle x=2,x=4$. Then $\displaystyle f(2) = 2\log(2) - (\log 2)^2>0$ and $\displaystyle f(4) = 4\log(2) - (\log 4)^2>0$, so since $\displaystyle f(x)$ increases everywhere else, it must be greater than zero for all $\displaystyle x>1$.