I'm having trouble solving this exercise: (find the domain of this function) $$f(x)=\frac{\ln(4-x^2)}{\ln(x+1)}$$
Here is my attempt:
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I've found the domain of the natural logarithm in the numerator, which is equal to: $$4-x^2>0$$ (it must be strictly positive),
1.1 $$-x^2>-4$$
1.2 $$x^2<4$$
1.3 (result) $$-2<x<2$$
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I've found the domain of the natural logarithm in the denominator, which is equal to: $$x+1>0$$
1.1 (result) $$x>-1$$
Now, I have to find the interval (on the number line) in which each possible solution is acceptable. So, I've plotted each result on a number line, here is the result:
$$-1<x<2$$
but this question has these possible solutions (it is a multiple choice question):
(1) $-1<x<2, except\space0$
(2) $-\infty<x<-1$
(3) $-2<x<-1$
(4) $-\infty<x<0$
The solution is $(1)$, but I don't understand why is not 0 included in the solution's interval.
EDIT (correct answer):
In the denominator I would have had considered the equation involving the denominator: $\ln(x+1)=0$. it mustn't be equal to 0.
this error occurs because of distraction.
Best Answer
At $x=0$ we have $\ln (1-0) = \ln (1) = 0$ in the denominator, which is undefined.