Find minimal integer $n>1$ for which $2^n > n^{1000}$. Find the minimum integer $n>1$ for which $2^n > n^{1000}$.
I have taken the $log$ on both sides, but not reached any result. I would appreciate if anybody will solve it accurately. Thanks in advance.
 A: Hint:
You can look, equivalently, at the inequality $n > \frac{1000}{\ln 2} \ln n$.
Now, study the function $f\colon x> 0 \mapsto x - \frac{1000}{\ln 2} \ln x$ (with the usual tools: differentiation, etc.) to see starting at which value it is increasing. (It's not hard to see that $f$ is decreasing, then increasing). This will help you figure out the minimum value $x_0$ such that $f(x) > 0$ for all $x> x_0$.
A: We have n ≥ 2. Therefore $2^n > 2^{1000}$, which makes n ≥ 1,001. 
We have n ≥ 1,001. Therefore $2^n > 1001^{1000}$, which makes n ≥ 9,967.226 or n ≥ 9,968. 
We have n ≥ 9,968. Therefore $2^n > 9968^{1000}$, which makes n ≥ 13,283.088 or n ≥ 13,284.  
We have n ≥ 13,284. Therefore $2^n > 13284^{1000}$, which makes n ≥ 13,697.402 or n ≥ 13,698.  
We have n ≥ 13,698. Therefore $2^n > 13698^{1000}$, which makes n ≥ 13,741.677 or n ≥ 13,742.  
We have n ≥ 13,742. Therefore $2^n > 13742^{1000}$, which makes n ≥ 13,746.303 or n ≥ 13,747.  
Now $log_2 (13,747) = 13,746.829$, which makes $2^{13,747} > 13,747^{1,000}$. 
A: Taking $\log$ on both sides, you get
$$n \log 2 > 1000 \log n$$
or $$n - \frac{1000}{\log 2} \log n >0$$
This resembles the upper left triangular portion of the real plane formed by the straight line $$n - \frac{1000}{\log 2} \log n = 0$$
Hope this helps.
