4
$\begingroup$

From other posts on here I learned that the following is true:

$$ \log(a + b) = \log\left(a \cdot \left(1 + \frac{b}{a}\right)\right) = \log(a) + \log\left(1 + \frac{b}{a}\right) $$

What about $\log(a + b + c)$?

$\endgroup$
  • 3
    $\begingroup$ $log(a+b+c)=log((a+b)+c)$ $\endgroup$ – ModOverRing May 1 '17 at 11:28
4
$\begingroup$

$$\log((a+b)+c)=\log\left((a+b)\cdot\left(1+\frac{c}{a+b}\right)\right)$$ $$=\log(a)+ \log\left(1+\frac{b}{a}\right)+\log\left(1+\frac{c}{a+b}\right)$$

$\endgroup$
  • $\begingroup$ @ClaudeLeibovici Sorry? I can't get you. $\endgroup$ – The Dead Legend May 1 '17 at 12:56
  • $\begingroup$ It was a + originally, idk why it was edit. @ClaudeLeibovici $\endgroup$ – The Dead Legend May 1 '17 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.