Question 550585
We have


*[tex \LARGE \log(n) \ge \log(2^k) \Rightarrow n \ge 2^k]


This is true because if we let *[tex \LARGE n = p_1^{a_1}p_2^{a_2}...p_k^{a_k}], then we have


*[tex \LARGE p_1^{a_1}p_2^{a_2}...p_k^{a_k} \ge 2*2*2*...*2], and since each prime is greater than or equal to 2, this inequality holds.