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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}

Final Answer


(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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