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Let´s say we have the following function:

$$f(x)= \log\left(\frac{x + 3}{x-3}\right)$$

Now we are looking for the domain of $f(x)$. The argument has to be strictly positive. The are two cases:

1) Numerator and denominator are strictly positive.

$x+3>0 \cap x-3>0 \Rightarrow x>3$

2) Numerator and denominator are strictly negative.

$x+3<0 \cap x-3<0 \Rightarrow x<-3$

The union is $x\in(-\infty,-3)\cup(3,\infty)$. So far so good. Now we can write $f(x)$ in a different way by using logarithm rules.

$$f(x)= \log(x + 3)-\log(x-3)$$

In this case $f(x)$ is not defined for $x<-3$, since both arguments are negative.

My Question: How can this (apparent) contradiction be resolved?

Thanks for reading my question.

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There's no contradiction: in fact $\log(ab)=\log(a)+\log(b)$ if and only if $a,b>0$.

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  • $\begingroup$ Ok. But how do I know which is the right way of expressing $f(x)$? $\endgroup$ – callculus Oct 30 '19 at 18:46
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    $\begingroup$ The "right way" is the form your expression is given. In particular, in the first way your function is defined for $x<-3\vee x>3$ as you pointed out, while in the second way the function is defined only for $x>3$, and in this range is equal to the first one. $\endgroup$ – Lorenzo Cecchi Oct 30 '19 at 18:48
  • $\begingroup$ So they are different functions? $\endgroup$ – callculus Oct 30 '19 at 18:49
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    $\begingroup$ Definitely yes -- although they're the same one for $x>3$. $\endgroup$ – Lorenzo Cecchi Oct 30 '19 at 18:51
  • $\begingroup$ OK, thank you very much for explanation. $\endgroup$ – callculus Oct 30 '19 at 18:52
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Adding to the answer above, which correctly states that the logarithmic property $\log (xy) = \log x + \log y$ holds iff $x,y > 0$ :

If you are given the expression $f(x) = \log\left( \frac{x+3}{x-3}\right)$ then your argument is $\frac{x+3}{x-3}$ and you need to work with the domain restrictions applying to that. Furthermore, applying the property will only be viable for the corresponding parts of the domain.

This means, that $f(x) = \log\left(\frac{x+3}{x-3}\right)$ and $g(x) = \log(x+3) - \log(x-3)$ are two different functions.

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    $\begingroup$ Thanks for your answer. It confirms what Lorenzo has written. $\endgroup$ – callculus Oct 30 '19 at 18:54

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