1. **State the problem:** Solve the inequality $\log_2(3-x) > 0$. 2. **Recall the logarithm rule:** For $\log_b(A) > 0$ where $b > 1$, this means $A > 1$ because $\log_b(1) = 0$. 3. **Apply the rule:** Since the base is 2 (which is greater than 1), we have: $$3 - x > 1$$ 4. **Solve the inequality:** $$3 - x > 1$$ $$-x > 1 - 3$$ $$-x > -2$$ Multiply both sides by $-1$ (remember to reverse inequality): $$x < 2$$ 5. **Domain restriction:** The argument of the logarithm must be positive: $$3 - x > 0$$ $$x < 3$$ 6. **Combine conditions:** $$x < 2$$ and $$x < 3$$ The more restrictive condition is $$x < 2$$. **Final solution:** $$\boxed{x < 2}$$