1. **State the problem:** Solve for $x$ in the equation $$\log_4(x - 2) + \log_4(x - 2) = 0.$$\n\n2. **Use the logarithm property:** When adding logarithms with the same base, we multiply the arguments: $$\log_4(x - 2) + \log_4(x - 2) = \log_4\big((x - 2)(x - 2)\big) = \log_4\big((x - 2)^2\big).$$\n\n3. **Rewrite the equation:** The equation becomes $$\log_4\big((x - 2)^2\big) = 0.$$\n\n4. **Convert logarithmic to exponential form:** Recall that $$\log_b(a) = c \iff b^c = a.$$ Applying this, we get $$4^0 = (x - 2)^2.$$\n\n5. **Simplify:** Since $$4^0 = 1,$$ we have $$1 = (x - 2)^2.$$\n\n6. **Solve the quadratic equation:** Taking the square root of both sides, $$\sqrt{1} = \sqrt{(x - 2)^2} \Rightarrow \pm 1 = x - 2.$$\n\n7. **Find the values of $x$:**\n- For $x - 2 = 1$, $$x = 3.$$\n- For $x - 2 = -1$, $$x = 1.$$\n\n8. **Check the domain:** The argument of the logarithm must be positive: $$x - 2 > 0 \Rightarrow x > 2.$$\nOnly $x = 3$ satisfies this condition.\n\n**Final answer:** $$x = 3.$$