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Let $K \geq1$ be an arbitrary positive constant. Why do we have the equality $$\sum_{x/K \leq p \leq x} \dfrac{log(p)}{p} = log(K)+O(1),$$ where the error term is uniform in $K?$ Here the summation is over the primes in the given interval.

inequalitynoob2
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    follows immediately from $\sum_{p \le N} \frac{\log(p)}{p} = \log(N)+O(1)$, a proof of which you can find here. https://math.stackexchange.com/questions/33980/how-to-prove-chebyshevs-result-sum-p-leq-n-frac-log-pp-sim-log-n-a – mathworker21 Aug 15 '19 at 14:20
  • an intuitive proof that $\sum_{p \le N} \frac{\log(p)}{p} = \log(N)+O(1)$ is that the sum is basically $\sum_{n \le N} \frac{\log(n)}{n}\frac{1}{\log(n)}$ (where we used that the "probability" an integer $n$ is prime is roughly $\frac{1}{\log n}$), which gives $\sum_{n \le N} \frac{1}{n} = \log(N)+O(1)$. – mathworker21 Aug 15 '19 at 14:21
  • @mathworker21 Is there a way to get this without using the result of Chebyshev you mentioned? – inequalitynoob2 Aug 15 '19 at 14:23
  • you mean that $\sum_{p \le N} \frac{\log p}{p} = \log(N)+O(1)$????? that's a special case of your question, since you want uniformity in $K$. its equivalent to your question of course – mathworker21 Aug 15 '19 at 14:58

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