Question

If "n" is a natural number such that $n=p_1^{\alpha 1}.p_2^{\alpha 2}...............p{k}^{\alpha 2};p_1,p_2..............,p_k$ are distinct primes, then

• A. log $n\ge k$ log 1.
• B. log $n\ge k$ log 2.
• C. log $n\ge k$ log 3.
• D. log $n\ge 2k$ log 2.

Answer 1

Answer: (B)

log $n\ge k$ log 2.

$\log \left(n\right)={\alpha }_1\log p_1+{\alpha }_2\log p_2+....{\alpha }_k\log p_k$

Since n is a natural number, $p_1,p_2.....p_k$ are primes, ${\alpha }_1........{\alpha }_k$ must be natural numbers.

Also as $p_i\ge 2,{\alpha }_i\ge 1$, we get

$\log n\ge \log 2+\log 2+\log 2+...\left(ktimes\right)=k\times \log 2$