1. **State the problem:** Solve the inequality $\log_{\frac{1}{2}}(x-3) > 1$. 2. **Recall the logarithm inequality rule:** For a base $b$ where $0 < b < 1$, the logarithm function is decreasing. This means if $\log_b(A) > C$, then $A < b^C$. 3. **Apply the rule:** Here, the base is $\frac{1}{2}$ which is between 0 and 1, so the inequality reverses: $$x - 3 < \left(\frac{1}{2}\right)^1$$ 4. **Simplify the right side:** $$x - 3 < \frac{1}{2}$$ 5. **Solve for $x$:** $$x < 3 + \frac{1}{2}$$ $$x < \frac{7}{2}$$ 6. **Domain restriction:** Since the argument of the logarithm must be positive: $$x - 3 > 0 \implies x > 3$$ 7. **Combine domain and inequality:** $$3 < x < \frac{7}{2}$$ **Final answer:** $$\boxed{3 < x < \frac{7}{2}}$$