1. **State the problem:** Solve the equation $ (x^2 - 3) \times \ln(3x) = 0 $. 2. **Formula and rules:** For a product of two factors to be zero, at least one of the factors must be zero. So, we set each factor equal to zero: $$ x^2 - 3 = 0 \quad \text{or} \quad \ln(3x) = 0 $$ 3. **Solve the first factor:** $$ x^2 - 3 = 0 $$ Add 3 to both sides: $$ x^2 = 3 $$ Take the square root of both sides: $$ x = \pm \sqrt{3} $$ 4. **Solve the second factor:** $$ \ln(3x) = 0 $$ Recall that $\ln(a) = 0$ means $a = 1$, so: $$ 3x = 1 $$ Divide both sides by 3: $$ x = \frac{1}{3} $$ 5. **Check domain restrictions:** The argument of the logarithm must be positive: $$ 3x > 0 \implies x > 0 $$ So, $x = -\sqrt{3}$ is not valid because it makes $3x$ negative. 6. **Final solution:** $$ x = \sqrt{3} \quad \text{or} \quad x = \frac{1}{3} $$ These are the values of $x$ that satisfy the original equation.