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Is there a way to express $\ln (x+1)$ as $A\ln x + B$, where $A$ and $B$ are some expressions in $x$ that do not themselves include logarithmic terms, $A\neq 0$, and $x>1$?

Shophaune
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    How about $A=0$ and $B=\ln(x+1)$? – Karl Jul 02 '23 at 14:28
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    Yes. In fact, your system is over-determined. Let $B=-A ln(x) + ln(x+1)$ and $A$ be arbitrary. I know that's not what you had in mind, but it's true. – Dr. Momo Jul 02 '23 at 14:32
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    @Karl I assume the OP meant $A \neq 0$, $B\neq 0$, even if he did not write it explicitly. Or... I like to think it was understood, otherwise the question is rather useless. – Enrico M. Jul 02 '23 at 14:42
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    @EnricoM. Without more details about $A$ and $B$ the question is still very vague: Take $A=\frac{\sin x}{\ln(x)}, B=\ln(x+1)-\sin(x)$ – jjagmath Jul 02 '23 at 14:45
  • @jjagmath Indeed, there are infinite possibilities... The OP is invited to add some details! – Enrico M. Jul 02 '23 at 14:49
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    Clarified what I was looking for...I hope. If it's still too vague do say! – Shophaune Jul 02 '23 at 14:52
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    I think this is still too vague. For example, do you consider $\ln(x)+\ln(1+\frac{1}{x})$ a valid answer? – jjagmath Jul 02 '23 at 14:56
  • It's certainly more valid, but still not quite what I'm looking for - because then my next question would be expressing $\ln(1+\frac{1}{x})$ in terms of $\ln(x)$ which would just come right back here. Localth's answer seems to be what I was looking for though, provided there's nothing factually wrong with it that I'm unaware of?

    Edit - it's been deleted so I assume there was.

     – Shophaune Jul 02 '23 at 15:06
  • @Shophaune The answer you thought to be good is just the same answer jjagmath gave you, for $$\sum_{k = 1}^{+\infty} \dfrac{(-1)^k x^{-k}}{k!} = - \ln\left(\dfrac{x+1}{x}\right) \equiv \ln(x+1) - \ln(x)$$

    If you only take few terms of the series, then you are making an approximation, so no equality holds.

     – Enrico M. Jul 02 '23 at 15:09
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    I had no idea about that series when posting the question nor when jjagmath posted their comment, but regardless it's exactly what I was looking for. – Shophaune Jul 02 '23 at 15:12
  • @EnricoM. Oh, actually, I should ask - the series you've posted differed from Localth's answer with the inclusion of a factorial, which form is correct? – Shophaune Jul 02 '23 at 15:16
  • @Shophaune You're right, my bad! There is NO factorial :) – Enrico M. Jul 03 '23 at 05:50

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$$\ln(x+1)=\ln(x)-\sum_{k=1}^\infty\frac{(-1)^kx^{-k}}{k}, \text{for $|x|>1$}$$

Localth
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