Correct Answer - A
We have,
`n = p_(1)^(a1) * p_(2)^(a2) * p_(3)^(a3) …..p_(k)^(ak)`
Since `"Pi"^(rs)` are primes and n is a natural number. Therefore, each `a_(1)` is a natural number.
Also, `"pi" ge 2 "for" I = 1, 2 ,…. k " " [because 2 "is the smallest prime"]`
Now, ` n=p_(1)^(a1) * p_(2)^(a2) * p_(3)^(a3)... p_(k)^(ak)`
`rArr "log"n = a_(1) "log" p_(1) +a_(2) "log" p_(2) +... + a_(k) "log" p_(k)`
`rArr "log"n ge a_(1)"log" 2 +a_(2) "log" 2 + .... + a_(k) "log" 2`
`rArr "log" n ge (a_(1) + a_(2) +.... +a_(k))"log" 2`
`rArr "log" n ge k"log" 2 " " [{:(because , a_(1) ge 1 "for" i = 1"," 2",...," k),(therefore, a_(1) + a_(2) +... + a_(k) ge k):}]`