Find the Domain y = log of 9x+1

$y = \log (9 x + 1)$

Set the argument in $\log (9 x + 1)$ greater than $0$ to find where the expression is defined.

$9 x + 1 > 0$

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$9 x > - 1$

Divide each term in $9 x > - 1$ by $9$ and simplify.

Divide each term in $9 x > - 1$ by $9$ .

$\frac{9 x}{9} > \frac{- 1}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x}{9} > \frac{- 1}{9}$

Divide $x$ by $1$ .

$x > \frac{- 1}{9}$

Simplify the right side.

Move the negative in front of the fraction.

$x > - \frac{1}{9}$

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

Interval Notation:

$(- \frac{1}{9}, \infty)$

Set-Builder Notation:

${x | x > - \frac{1}{9}}$