Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have two numbers $x$ and $y$ which are $\log_2$ of two other numbers $P$ and $V$. I'm trying to calculate $\log_2 (P-V)$ without transforming $x$ and $y$ back to the linear world.

If $P = AV$, then $P-V = V(A-1)$, and $x-y = \log_2 A$. So, if I could calculate $\log_2 (A-1)$ I could just add it to $y$ and be done. Is there a relationship I can use to do this?

share|cite|improve this question
Do you mean $\log(A-1)$ or $\log A -1$? – Asaf Karagila Dec 22 '10 at 10:10
@Asaf - log(A-1). I fixed the question, thanks. – mtrw Dec 22 '10 at 10:15

One way is to write (am using natural logs for now): $\log(A-1) = \log A + \log (1-1/A)$, and to then use the power series for $\log(1-x)$ to approximate the latter term. But this is not really a good solution.

share|cite|improve this answer
It's not that bad. If $A$ is large, the series should converge quite rapidly. If $A$ is smaller than $2$, then it may be smarter to change tactic and use a different approximation: $\log(A-1)=\log(1+(A-2))$ and then use the series expansion. – Raskolnikov Dec 22 '10 at 12:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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