# how to find the domain of this function:$f(x)=\frac{\ln(4-x^2)}{\ln(x+1)}$

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

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:

$$-1

but this question has these possible solutions (it is a multiple choice question):

(1) $$-1

(2) $$-\infty

(3) $$-2

(4) $$-\infty

The solution is $$(1)$$, but I don't understand why is not 0 included in the solution's interval.

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.
• Notice thta $\ln(x+1)=0\iff x=0$, so you must avoid null denominators. Dec 12 '20 at 9:17
At $$x=0$$ we have $$\ln (1-0) = \ln (1) = 0$$ in the denominator, which is undefined.