1. Problem: Find the product of the roots of the equation $$(\log_3 x)^2 - 3 \log_3 x - 4 = 0.$$\n\n2. Step 1: Let $y = \log_3 x$. Then the equation becomes $$y^2 - 3y - 4 = 0.$$\n\n3. Step 2: This is a quadratic equation in $y$. The product of the roots of $ay^2 + by + c = 0$ is given by $$\frac{c}{a}.$$\nHere, $a=1$, $b=-3$, and $c=-4$.\n\n4. Step 3: Calculate the product of the roots: $$\frac{c}{a} = \frac{-4}{1} = -4.$$\n\n5. Step 4: Since $y = \log_3 x$, the roots $y_1$ and $y_2$ correspond to $x_1 = 3^{y_1}$ and $x_2 = 3^{y_2}$.\n\n6. Step 5: The product of the roots in terms of $x$ is $$x_1 \cdot x_2 = 3^{y_1} \cdot 3^{y_2} = 3^{y_1 + y_2}.$$\n\n7. Step 6: The sum of the roots of the quadratic is $$-\frac{b}{a} = -\frac{-3}{1} = 3.$$\n\n8. Step 7: Therefore, $$x_1 \cdot x_2 = 3^{3} = 27.$$\n\nAnswer: The product of the roots of the original equation is **27**.