1. **State the problem:** Find the domain of the function $$f(x) = \ln(x^2 - 12x)$$. 2. **Recall the domain condition for natural logarithm:** For $$f(x) = \ln(u)$$, the argument $$u$$ must satisfy $$u > 0$$. 3. **Apply this to our function:** We need $$x^2 - 12x > 0$$. 4. **Factor the inequality:** $$x^2 - 12x = x(x - 12) > 0$$. 5. **Solve the inequality:** For the product to be positive, either both factors are positive or both are negative. - Case 1: $$x > 0$$ and $$x - 12 > 0 \implies x > 12$$ - Case 2: $$x < 0$$ and $$x - 12 < 0 \implies x < 0$$ 6. **Domain from inequality:** $$x < 0$$ or $$x > 12$$. 7. **Write the domain in interval notation:** $$(-\infty, 0) \cup (12, \infty)$$. **Final answer:** The domain of $$f(x) = \ln(x^2 - 12x)$$ is $$(-\infty, 0) \cup (12, \infty)$$.