Find the Domain y = log of 8+3x

$y = \log (8 + 3 x)$

Set the argument in $\log (8 + 3 x)$ greater than $0$ to find where the expression is defined.

$8 + 3 x > 0$

Solve for $x$ .

Subtract $8$ from both sides of the inequality.

$3 x > - 8$

Divide each term in $3 x > - 8$ by $3$ and simplify.

Divide each term in $3 x > - 8$ by $3$ .

$\frac{3 x}{3} > \frac{- 8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} > \frac{- 8}{3}$

Divide $x$ by $1$ .

$x > \frac{- 8}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x > - \frac{8}{3}$

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

Interval Notation:

$(- \frac{8}{3}, \infty)$

Set-Builder Notation:

${x | x > - \frac{8}{3}}$