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We have a logarithmic function $f(x) = \log_3[(x-2)(x-3)]$. In order to determine a domain of this function we have to solve an equation $(x - 2)(x - 3) > 0$. The result is a range $(-\infty, 2) \cup (3, +\infty)$.

But now we can transform this function to the following form: $f(x) = log_3(x - 2) + log_3(x - 3)$ and now a domain is a range $(3, +\infty)$.

I don't understand why after transformation the domain of function f is different. Shouldn't it be the same? Is there a mistake in my reasoning or maybe I should find a domain after every transformation? Or maybe in some cases such a transformation is disallowed?

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Your transformation is not legal, if you don't know the signs of the individual factors. $\log(ab)=\log(a)+\log(b)$ only if $a>0$ and $b>0$. If $a<0$ and $b<0$, the legal transformation is $\log(ab)=\log(-a)+\log(-b)$.

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The $log_3[(x-2)(x-3)]$ is defined in the cases $(x-2)>0$, $(x-3)>0$ and $(x-2)<0$, $(x-3)<0$.

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