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I know for a fact that it is not $\log x + \log y$, but Im unsure as to how to proceed.. I have checked the basic log properties but nowhere do they give an example of a statement like the one above.

Thanks in advance

Jeel Shah
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Peter
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  • cause it has no standard expansion, at least not a usefull one, for what do you need one ? – Dominic Michaelis Apr 02 '13 at 11:01
  • There no such formula for $log(x+y)$ . But you can use the logarithmic series expansion if needed. – lsp Apr 02 '13 at 11:02
  • http://math.stackexchange.com/questions/348956/how-to-solve-the-following-recurrence-tn-leq-2c-lfloor-fracn2-rfloor@DominicMichaelis, refer to this post and you'll see why I need it. – Peter Apr 02 '13 at 11:04
  • There is nothing pleasant. If $y\ne 0$, we can rewrite $x+y$ as $y(1+\frac{x}{y})$, and then get $\log(x+y)=\log y +\log\left(1+\frac{x}{y}\right)$. It is possible to imagine a situation where this could be useful. – André Nicolas Apr 02 '13 at 11:04
  • You could expand using the Taylor series... – pshmath0 Oct 17 '13 at 07:24
  • $$ \log(x+y)=\log(x)+\log(y)+\log\bigg(\frac{1}{y}+\frac{1}{x}\bigg) $$ or $$ \log(x+y)=h \log(x)+k \log(y)+\log\bigg(\frac{1}{y^{k} x^{h-1}}+\frac{1}{x^{h} y^{k-1}}\bigg) $$ – Patrick Danzi Sep 25 '20 at 11:22

1 Answers1

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There's no easy, or more or less generally useful, expansion of this thing. You could try though:

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

The above is assuming $\,x>0\,$, and it doesn't look that nice or useful, does it? But who knows, perhaps under certain circumstances...

Of course, you can do the above factoring out $\,y\,$, too.

DonAntonio
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