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