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if n is a natural number such that n=p1 power a1 *p2 power a2*p3 power a3..............*pk power ak and pk are distinct primes , then show that log n>=k log2
n= p(1)^a(1) + p(2)^a(2) + p(3)^a(3) + .........+ p(k)^a(k).n is a natural number,,,this leads to n=1 or larger integer. ,,,, AND it limits each term of the sequence to a positive number. That is, the p"s cannot be + or -, as this would restrict the terms to an even number, to make n positive, which was not stated..Likewise, the "a" values cannot be negative, leading to a fraction, or a fraction, which would lead to not whole numbers..The given that p(k) is a distinct prime, leads to the min value of p(k) being the min prime, or 2 .,,,or larger,,, (1 is a special number, not prime).n = something positive * a number 2 or larger raised to a positive number( making it >2).n>2^ k.and taking log of both sides.log n > log 2^k or log n > k log 2 Answer By Suyash Mishra Kanpur
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