1. **State the problem:** Find the value of $x$ from the equation $\log_2 x + \log_2 (x-3)$.\n\n2. **Use logarithm properties:** Recall that $\log_a m + \log_a n = \log_a (m \times n)$.\nSo, $\log_2 x + \log_2 (x-3) = \log_2 [x(x-3)] = \log_2 (x^2 - 3x)$.\n\n3. **Equation form:** For the expression to be equal to a number (usually zero if not otherwise specified), we consider $\log_2 (x^2 - 3x) = 0$.\n\n4. **Solve the equation:** $\log_2 (x^2 - 3x) = 0 \implies x^2 - 3x = 2^0 = 1$.\n\n5. **Form quadratic:** $x^2 - 3x - 1 = 0$.\n\n6. **Use quadratic formula:** For $ax^2 + bx + c = 0$, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=1$, $b=-3$, $c=-1$.\n$$x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2} = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}$$\n\n7. **Check domain restrictions:** Since the original logs require $x > 0$ and $x - 3 > 0 \Rightarrow x > 3$, only values greater than 3 are valid.\n\n8. **Evaluate values:**\n- $\frac{3 + \sqrt{13}}{2} \approx \frac{3 + 3.606}{2} = \frac{6.606}{2} = 3.303$, which is $> 3$, valid.\n- $\frac{3 - \sqrt{13}}{2} \approx \frac{3 - 3.606}{2} = \frac{-0.606}{2} = -0.303$, not valid.\n\n**Final answer:** $$x = \frac{3 + \sqrt{13}}{2}$$