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

The domain of f(x)= 1/(2+ln(x)) is 0 1/(e^2)

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domain of f(x)= 1/(2+ln(x))

domain

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

0 < x < \frac{1}{e ^{2}} or x > \frac{1}{e ^{2}}

Interval Notation

(0, \frac{1}{e ^{2}}) \cup (\frac{1}{e ^{2}}, \infty)

Solution steps

Find positive values for logs:

x > 0

Find undefined (singularity) points:

x = \frac{1}{e ^{2}}

Combine real regions and undefined points for final function domain

0 < x < \frac{1}{e ^{2}} or x > \frac{1}{e ^{2}}

y = \frac{x ^{2} - 36}{x + 1}

g (x) = lo g_{3} (8 - 2 x) + 1

f (x) = \frac{2 x + 4}{24 - 8 x}

f (x) = \frac{x + 2}{- x}

\frac{x ^{2} - 4}{9 x + 4}

What is the domain of f(x)= 1/(2+ln(x)) ?
The domain of f(x)= 1/(2+ln(x)) is 0<x< 1/(e^2)\lor x> 1/(e^2)