1. **State the problem:** We need to find all values of $x$ such that $$f(x) = \log_2 x + \log_2 (x - 2) = 3.$$ 2. **Recall the logarithm property:** The sum of logarithms with the same base can be combined as the logarithm of the product: $$\log_2 a + \log_2 b = \log_2 (a \times b).$$ 3. **Apply the property:** $$f(x) = \log_2 (x (x - 2)) = 3.$$ 4. **Rewrite the equation in exponential form:** Since $\log_2 y = 3$ means $y = 2^3$, we have $$x (x - 2) = 2^3 = 8.$$ 5. **Form a quadratic equation:** $$x^2 - 2x = 8.$$ 6. **Bring all terms to one side:** $$x^2 - 2x - 8 = 0.$$ 7. **Solve the quadratic equation:** Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-2$, $c=-8$: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-8)}}{2} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}.$$ 8. **Calculate the roots:** - $x = \frac{2 + 6}{2} = \frac{8}{2} = 4$ - $x = \frac{2 - 6}{2} = \frac{-4}{2} = -2$ 9. **Check domain restrictions:** The arguments of the logarithms must be positive: - $x > 0$ - $x - 2 > 0 \Rightarrow x > 2$ Only $x=4$ satisfies both conditions. $x=-2$ is invalid because $\log_2(-2)$ is undefined. **Final answer:** $x = 4$ only.