If n is a natural number such that n = p_1^_1 p_2^_2 p_3^_3 p_k^_k and p_1,p_2,.p_k are distinct primes, then show that logn klog2.

Given that n=p_1^_1 p_2^_2 p_3^_3 . p_k^_k n N and p_1, p_2, p_3, , p_k are distinct prime numbers on both sides of (1), we get n=_1 p_1+_2 p_2+ .+_k p_k every prime number is such that p_i 2 _e p_i _e 2 i=1, k (2) and (3) we getaligned& n _1 2+_2 2+_3 2+ .+_k 2 \\& n (_1+_2+ .+_k) 2 \\& _1+_2+ .+_k>k \\& (_1+_2++_n) 2> 2 \\& n k 2 . (By transitive

If n is a natural number such that n = p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot p_{3}^{\alpha_{3}} p_{k}^{\alpha_{k}} and p_{1},p_{2},.p_{k} are distinct primes, then show that logn ? klog2.: Sequence & Series (Mathematics)

Given that

n=p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdot p_3^{\alpha_3} \ldots . p_k^{\alpha_k}

\nwhere

n \in N

and

p_1, p_2, p_3, \ldots \ldots, p_k

are distinct prime numbers\nTaking

\log

on both sides of (1), we get\n

\n\log n=\alpha_1 \log p_1+\alpha_2 \log p_2+\cdots .+\alpha_k \log p_k\n

\n \nSince every prime number is such that

p_i \geq 2

\n\therefore \log _e p_i \geq \log _e 2\n

\n\forall i=1, \ldots k\n

\n \nUsing (2) and (3) we get\n

\n\begin{aligned}\n& \log n \geq \alpha_1 \log 2+\alpha_2 \log 2+\alpha_3 \log 2+\cdots .+\alpha_k \log 2 \\\n& \Rightarrow \log n \geq\left(\alpha_1+\alpha_2+\cdots .+\alpha_k\right) \log 2 \\\n& \Rightarrow \alpha_1+\alpha_2+\cdots .+\alpha_k>k \\\n& \therefore\left(\alpha_1+\alpha_2+\cdots+\alpha_n\right) \log 2>\log 2 \\\n& \Rightarrow \log n \geq k \log 2 . \text { (By transitive law) }\n\end{aligned}\n

Explanation

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