What is the domain of y=log_{2}(((x+1))/(x^2-9)) ?

The domain of y=log_{2}(((x+1))/(x^2-9)) is -3 3

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domain of y=log_{2}(((x+1))/(x^2-9))

domain

y = lo g_{2} (\frac{( x + 1 )}{x ^{2} - 9})

- 3 < x < - 1 or x > 3

Interval Notation

(- 3, - 1) \cup (3, \infty)

Solution steps

Find positive values for logs:

- 3 < x < - 1 or x > 3

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

- 3 < x < - 1 or x > 3

- 3 sin (3 x)

2 \cdot lo g_{2 x + 3} (x)

y = sin (x^{2})

x + 3 + 3 - x

What is the domain of y=log_{2}(((x+1))/(x^2-9)) ?
The domain of y=log_{2}(((x+1))/(x^2-9)) is -3<x<-1\lor x>3