If n is a natural number such than $n=p_1^{a_1}.p_2^{a_2}.p_3^{a_3}.......p_k^{a_k}$ and $p_1,p_2,...p_k$ are distinct primes, then show that $log(n)>=klog(2)$.

What I think is p1 is greater than or equal to 2. So is p2 and all the way to pn. So their product should be greater than and equal to $2^{k}$. Is this proof sufficient?

Source: IIT JEE 1984

  • 1
    $\begingroup$ Yes, it is correct. $\endgroup$
    – TBTD
    Apr 23, 2017 at 19:04
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    $\begingroup$ Yes you're right. $\endgroup$
    – Iti Shree
    Apr 23, 2017 at 19:20
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    $\begingroup$ Yes.It's right... If you write p_1 then iwith dollar signs you get $p_1$ which is better than $p1$. ... Click on Help at the top bar, choose the Help Center, choose the Q "How can I format mathematics here?" $\endgroup$ Apr 23, 2017 at 23:04
  • $\begingroup$ @user254665 Thank you. i did check the post, but it had very few examples. $\endgroup$ Apr 24, 2017 at 3:09
  • $\begingroup$ Did you go to the bottom line and click on the MathJax quick reference? $\endgroup$ Apr 25, 2017 at 2:28


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