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

The domain of f(x)=(ln(x+1))/(e^x-1) is -1 0

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

domain

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

- 1 < x < 0 or x > 0

Interval Notation

(- 1, 0) \cup (0, \infty)

Solution steps

Find positive values for logs:

x > - 1

Find undefined (singularity) points:

x = 0

Combine real regions and undefined points for final function domain

- 1 < x < 0 or x > 0

f (x) = x 6 - x

4 (4 x^{2} - 3) - 3

\frac{4 x}{5 x - 6}

f (x) = 7 x - 3

f (x) = \frac{x}{x ^{2} - 4 x - 12}

What is the domain of f(x)=(ln(x+1))/(e^x-1) ?
The domain of f(x)=(ln(x+1))/(e^x-1) is -1<x<0\lor x>0