Question

If $n$ is a natural number such that $n=p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3}....p_k^{a_k}$ where $p_1,p_2,p_3,...,p_x$ are distinct prime numbers, then show that $\log n>k\log 2$.

Answer 1

$\log n=\log (p_1^{a_1}.p_2^{a_2}.......p_k^{a_k})$

$\log n=a_1\log p_1+a_2\log p_2+...........a_k\log p_k$

Now as all prime numbers are distinct and we also know that minimum prime number can be $2$ and then all will be greater than that, So

$\log n>a_1\log 2+a_2\log 2+...........a_k\log 2$

$\log n>(a_1+a_2+.....+a_k)\log 2$

Know $a_1,a_2,...a_k\ge 1$,

$\therefore \log n>(1+1+...k$times$)\log 2$

$\Rightarrow \log n>k\log 2$

Hence proved.