What is the domain of f(x)= 1/(ln(x^2-1)) ?

The domain of f(x)= 1/(ln(x^2-1)) is x sqrt(2)

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domain of f(x)= 1/(ln(x^2-1))

domain

f (x) = \frac{1}{ln ( x ^{2} - 1 )}

x < - 2 or - 2 < x < - 1 or 1 < x < 2 or x > 2

Interval Notation

(- \infty, - 2) \cup (- 2, - 1) \cup (1, 2) \cup (2, \infty)

Solution steps

Find positive values for logs:

x < - 1 or x > 1

Find undefined (singularity) points:

x = 2, x = - 2

Combine real regions and undefined points for final function domain

x < - 2 or - 2 < x < - 1 or 1 < x < 2 or x > 2

f (x) = (5 - e^{x})^{2}

5^{x}

f (x) = \frac{15}{x ^{2} + 5 x}

f (x) = \frac{x ^{5} - 1}{3 x ^{2} - 2 x - 1}

g (x) = 3 x - 1 + 5

What is the domain of f(x)= 1/(ln(x^2-1)) ?
The domain of f(x)= 1/(ln(x^2-1)) is x<-sqrt(2)\lor-sqrt(2)<x<-1\lor 1<x<sqrt(2)\lor x>sqrt(2)