If $n$ is a natural number such that $n={p_1}^{a_1}{p_2}^{a_2}\cdots \cdots {p_k}^{a_k},p_1,p_2\cdots \cdots p_k$ are distinct primes, then  ${\log }_{2}n$is

Given  $n={p_1}^{a_1}{p_2}^{a_2}\cdots \cdots {p_k}^{a_k}$

${\log }_{2}n={\log }_2({p_1}^{a_1}.{p_2}^{a_2}\cdots \cdots {p_k}^{ak})$

$={\log }_2{p_1}^{a_1}+{\log }_2{p_2}^{a_2}+\cdots \cdots +{\log }_2{p_k}^{ak}$

$=a_1{\log }_2{p}_1+a_2{\log }_2{p}_2+\cdots \cdots +ak{{\log }_2}^{p_k}$

$=a_1{{\log }_2}^{p_1}+a_2{{\log }_2}^{p_2}+\cdots +a_k{{\log }_2}^{p_k}$

$=a_1(\ge 1)+a_2(\ge 1)+\cdots \cdots +a_k(\ge 1)$

$\ge 1+1+\cdots \cdots +1\left(ktimes\right)$

$\ge k$