Question

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

Answer 1

Since n is the natural number and $p_i^{\prime }s$ are primes . it follows that $a_i^{\prime }s$ are also natural numbers .
Taking logarithms of both sides of given relation we get
${{\sum }_{i=1}^k}^{\prime }{a}_{i}log{p}_i\ge {{\sum }_{i=1}^k}^{\prime }{a}_{i}log2$
[ $\because$ smallest prime number is 2 ]
$\ge klog2,sinceeacha\ge 1and{p}_i\ge 2$