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.


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

  • $\begingroup$ Ok. But how do I know which is the right way of expressing $f(x)$? $\endgroup$ – callculus Oct 30 '19 at 18:46
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ Thanks for your answer. It confirms what Lorenzo has written. $\endgroup$ – callculus Oct 30 '19 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.