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

The domain of y= 1/(2xsqrt(2+ln(x))) is x> 1/(e^2)

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domain of y= 1/(2xsqrt(2+ln(x)))

domain

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

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

Interval Notation

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

Solution steps

Find non-negative values for radicals:

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

Find undefined (singularity) points:

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

Combine real regions and undefined points for final function domain

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

a (t) = \frac{1}{2} tan (t)

h (x) = \frac{5 x}{x ^{2} - 7 x + 10}

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

f (x) = \frac{x - 3}{x ^{2} - 7 x + 10}

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

What is the domain of y= 1/(2xsqrt(2+ln(x))) ?
The domain of y= 1/(2xsqrt(2+ln(x))) is x> 1/(e^2)