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:

  1. 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) $$-2<x<2$$

  2. 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:


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.

  • 1
    $\begingroup$ Notice thta $\ln(x+1)=0\iff x=0$, so you must avoid null denominators. $\endgroup$ Dec 12 '20 at 9:17
  • 1
    $\begingroup$ ah, okay, so it was an error of distraction. Thank you $\endgroup$ Dec 12 '20 at 9:19

At $x=0$ we have $\ln (1-0) = \ln (1) = 0$ in the denominator, which is undefined.


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