# On the sum of the logarithms of primes.

Let $p$ be a prime and $x$ be an integer. It is known that $\sum_{p\leq x} \log p = O(x)$, and i think this is equivalent to the Prime Number Theorem.
As a mere prospective undergraduate with only a minimal understanding of analytic number theory, i'm curiously wondering if there exists an analogous formula for $\sum_{p\leq x} \log(p-1)$ ?

• In this context, I think that $\log(p-1)$ and $\log p$ are of the same size. – André Nicolas Jan 30 '16 at 1:01
• They differ by roughly $\sum \frac1p\approx\log\log x$ – Empy2 Jan 30 '16 at 1:11
• @Michael, any reference or proof for that result would be most appreciated. Thank you. – User1 Jan 30 '16 at 1:13
• After a little research on the question in Martin Klazar's Introdution To Number Theory, i found the same answer that MichaeI provided below, but i was left with another question again, which i have posted here math.stackexchange.com/q/1632689/305659. Please may you kindly consider it also. Thank you. – User1 Jan 30 '16 at 2:26

$$\log(p-1)=\log(p)+\log(1-\frac1p)\approx\log(p)-\frac1p$$