Find the Domain natural log of 1/(x+2)

$\ln (\frac{1}{x + 2})$

Set the argument in $\ln (\frac{1}{x + 2})$ greater than $0$ to find where the expression is defined.

$\frac{1}{x + 2} > 0$

Solve for $x$ .

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Find the domain of $\frac{1}{x + 2}$ .

Set the denominator in $\frac{1}{x + 2}$ equal to $0$ to find where the expression is undefined.

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 2) \cup (- 2, \infty)$

The solution consists of all of the true intervals.

$x > - 2$

Set the denominator in $\frac{1}{x + 2}$ equal to $0$ to find where the expression is undefined.

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- 2, \infty)$

Set-Builder Notation:

${x | x > - 2}$