# Domain of One Variable Function – A function with ln inside a fraction – Exercise 5478

Exercise

Determine the domain of the function:

$$f(x)=\frac{\ln (x^2)}{\ln^2 (x) -4}$$

$$(0,e^{-2}) \cup (e^{-2},e^2) \cup (e^2,\infty)$$

Solution

Let’s find the domain of the function:

$$f(x)=\frac{\ln (x^2)}{\ln^2 (x) -4}$$

There is a ln function, so we need the expressions inside the ln to be positive:

$$x^2>0\text{ and }x> 0$$

The two inequalities result

$$x>0$$

There is also a denominator, therefore the denominator must be different from zero. We check when it equals zero:

$$\ln^2 (x) -4= 0$$

$$\ln^2 (x)= 4$$

$$\ln (x)= \pm 2$$

We got two solutions. One solution,

$$\ln x=2$$

We use the logarithm definition and get

$$x=e^2$$

Second solution,

$$\ln x=-2$$

Again, we use the logarithm definition and get

$$x=e^{-2}$$

Since we require the denominator to be other than zero, we get that x cannot hold these values. That is,

$$x\neq e^2 , e^{-2}$$

In summary, the function domain is

$$(0,e^{-2}) \cup (e^{-2},e^2) \cup (e^2,\infty)$$

Note: The meaning of the sign:

$$\cup$$

is union (“or” relation).

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