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I'm working on a math problem which might be solvable if I can re-express $\log(N+x)$ as $\log(N) +$ 'something.

The problem I am having with the Taylor series expansion about $x=0$ is that it carries infinitely higher powers of $N$ in the terms of the expansion, see here

Do you know how I might expand $\log(N+x)$ as $\log(N) +$ something? Any advice/comments/suggestions would really go a long way as I'm quite stuck.

Thanks for taking the time to consider this.

-McMath.

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    $\begingroup$ $log(x+y)=log(x(1+\frac{y}{x}))=log(x)+log(1+\frac{y}{x})$, probably not what you want though $\endgroup$ – gd1035 Oct 3 '18 at 18:49
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I do not know if this is helpful for you in your situation but you could write

$$\log(x+y)=\log(x)+\log\left(1+\frac yx\right)$$

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    $\begingroup$ In terms of practical usage, many programming environments offer a function log1p(), defined as $\mathrm{log1p}(x) = \log(1+x)$. This offers better accuracy if $x$ is near $0$. $\endgroup$ – njuffa Oct 3 '18 at 18:54
  • $\begingroup$ Although I don't think the expansion you provided is helpful for my specific application, it does technically answer the question, and has practical merit as njuffa already said. I appreciate your answer. Best. $\endgroup$ – McMath Oct 3 '18 at 18:57
  • $\begingroup$ @McMath Are you searching for approximation for large arguments $N$ or what exactly do you have in your mind? Maybe add your special context to your question to make things clear :) $\endgroup$ – mrtaurho Oct 3 '18 at 19:01
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Well, if you want $\log(x + y) = log x + K$ then the only way to do that (assuming $x > 0$) is

$K = \log(x+y) - \log x = \log (\frac {x+y}x) = \log (1+ \frac yx)$.

or to put it another way:

$\log(x +y) = \log(x(1 + \frac yx)) = \log x + \log(1 + \frac yx)$.

or to put it a third wy:

$\log x + K = \log x*m$ where $\log m = K$. And $x*m = x +y$. So $K = \log (1+\frac yx)$.

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