If n is a natural number such thatn n = p_1^alpha_1 * p_2^alpha_2 * p_3^alpha_3 ... ... p_k^alpha_kn and p_1,p_2, ... .p_k are distinct primes, then show thatn logn >= klog2.

Given that n=p_1^alpha_1 * p_2^alpha_2 * p_3^alpha_3 ... . p_k^alpha_knwhere n in N and p_1, p_2, p_3, ... ... , p_k are distinct prime numbersnTaking log on both sides of (1), we getnnlog n=alpha_1 log p_1+alpha_2 log p_2+ * s .+alpha_k log p_knn nSince every prime number is such that p_i >= 2nnTherefore log _e p_i >= log _e 2nnnforall i=1, ... knn nUsing (2) and (3) we getnn n log n >= alpha_1 log 2+alpha_2 log 2+alpha_3 log 2+ * s .+alpha_k log 2 n => log n >= (alpha_1+alpha_2+ * s .+alpha_k) log 2 n => alpha_1+alpha_2+ * s .+alpha_k>k n Therefore (alpha_1+alpha_2+ * s+alpha_n) log 2>log 2 n => log n >= k log 2 . (By transitive law) n n

If n is a natural number such that n = p_{1}^{\alpha_... | 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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