To find the domain of the function $f(x) = \log(8 + 3x)$, we need to ensure that the argument of the logarithmic function is positive. Logarithms are only defined when their argument is greater than zero. Thus, we must solve the inequality:

$8 + 3x > 0$

Step-by-step solution:

1. Subtract 8 from both sides:

$3x > -8$

2. Divide both sides by 3:

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

Therefore, the domain of the function is all values of $x$ greater than $-\frac{8}{3}$.

Answer in interval notation:

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

Would you like more details on any step or further clarification?

Here are 5 related questions you might want to explore:

1. What is the domain of $f(x) = \log(x^2 + 1)$?
2. How do you find the domain of a function involving square roots and logarithms?
3. What is the range of $f(x) = \log(8 + 3x)$?
4. Can the logarithm of a negative number ever be defined?
5. How does changing the base of the logarithm affect the domain?

Tip: When working with logarithmic functions, always make sure the argument of the logarithm is strictly greater than zero to determine the domain.