1. **Problem:** Solve the equation using logarithmic laws: $$\log_4 x + \log_4 (x - 6) = 2$$ 2. **Formula and rules:** Recall the logarithmic product rule: $$\log_b A + \log_b B = \log_b (AB)$$ 3. **Apply the product rule:** $$\log_4 x + \log_4 (x - 6) = \log_4 [x(x - 6)] = 2$$ 4. **Rewrite the equation:** $$\log_4 [x(x - 6)] = 2$$ 5. **Convert logarithmic form to exponential form:** $$x(x - 6) = 4^2$$ 6. **Calculate the right side:** $$x^2 - 6x = 16$$ 7. **Bring all terms to one side:** $$x^2 - 6x - 16 = 0$$ 8. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-6$$, $$c=-16$$ 9. **Calculate the discriminant:** $$\sqrt{(-6)^2 - 4(1)(-16)} = \sqrt{36 + 64} = \sqrt{100} = 10$$ 10. **Find the roots:** $$x = \frac{6 \pm 10}{2}$$ 11. **Calculate each root:** - $$x = \frac{6 + 10}{2} = \frac{16}{2} = 8$$ - $$x = \frac{6 - 10}{2} = \frac{-4}{2} = -2$$ 12. **Check for domain restrictions:** Since $$\log_4 x$$ and $$\log_4 (x - 6)$$ require $$x > 0$$ and $$x - 6 > 0$$, we must have $$x > 6$$. 13. **Discard invalid root:** $$x = -2$$ is invalid because it is less than 6. 14. **Final answer:** $$\boxed{8}$$