What is the domain of y=ln(sqrt(x^2+1)) ?

The domain of y=ln(sqrt(x^2+1)) is -infinity <x<infinity

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domain of y=ln(sqrt(x^2+1))

domain

y = ln (x^{2} + 1)

- \infty < x < \infty

Interval Notation

(- \infty, \infty)

Solution steps

Find positive values for logs:

- \infty < x < \infty

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

f (b) = 2 lo g_{b} (5)

f (x) = \frac{x - 4}{3 x - 3}

f (x) = - 4 x - 4

f (x) = lo g_{23} (x)

y = - x^{2} + 14 x + 51

What is the domain of y=ln(sqrt(x^2+1)) ?
The domain of y=ln(sqrt(x^2+1)) is -infinity <x<infinity