# Given $\log_2 A$, is there an easy way to calculate $\log_2 (A-1)$?

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?

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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.
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