1. **State the problem:** Solve the logarithmic equation $$\log_2 (x - 2) + \log_2 (x - 3) = 0$$ for $$x$$. 2. **Use logarithm properties:** Recall the product rule for logarithms: $$\log_b A + \log_b B = \log_b (AB)$$. 3. **Apply the product rule:** $$\log_2 (x - 2) + \log_2 (x - 3) = \log_2 \big((x - 2)(x - 3)\big) = 0$$ 4. **Rewrite the equation in exponential form:** Since $$\log_2 y = 0$$ means $$y = 2^0 = 1$$, we have $$ (x - 2)(x - 3) = 1 $$ 5. **Expand the product:** $$ x^2 - 3x - 2x + 6 = 1 $$ $$ x^2 - 5x + 6 = 1 $$ 6. **Bring all terms to one side:** $$ x^2 - 5x + 6 - 1 = 0 $$ $$ x^2 - 5x + 5 = 0 $$ 7. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-5$$, $$c=5$$. 8. **Calculate the discriminant:** $$ \Delta = (-5)^2 - 4 \times 1 \times 5 = 25 - 20 = 5 $$ 9. **Find the roots:** $$ x = \frac{5 \pm \sqrt{5}}{2} $$ 10. **Check domain restrictions:** Since $$\log_2 (x-2)$$ and $$\log_2 (x-3)$$ require $$x-2 > 0$$ and $$x-3 > 0$$, we must have $$x > 3$$. 11. **Evaluate roots against domain:** $$ \frac{5 - \sqrt{5}}{2} \approx 1.38 < 3 $$ (reject) $$ \frac{5 + \sqrt{5}}{2} \approx 3.62 > 3 $$ (accept) **Final answer:** $$ x = \frac{5 + \sqrt{5}}{2} $$ **Regarding the multiple choice:** - The correct exponential form is (B) $$(x - 2)(x - 3) = 2^0$$. - The correct solution is (A) $$x = \frac{5 \pm \sqrt{5}}{2}$$, but only the positive root satisfies the domain.