1. **State the problem:** Solve the inequality $$\ln(x-1) + \ln(x-3) > \ln(3)$$ for $x$. 2. **Use logarithm properties:** Recall that $$\ln(a) + \ln(b) = \ln(ab)$$ for $a,b>0$. 3. **Rewrite the inequality:** $$\ln((x-1)(x-3)) > \ln(3)$$ 4. **Exponentiate both sides:** Since the natural logarithm function is increasing, the inequality holds if and only if $$(x-1)(x-3) > 3$$ 5. **Expand and simplify:** $$(x-1)(x-3) = x^2 - 4x + 3$$ So the inequality becomes $$x^2 - 4x + 3 > 3$$ 6. **Subtract 3 from both sides:** $$x^2 - 4x + 3 - 3 > 0 \implies x^2 - 4x > 0$$ 7. **Factor the quadratic:** $$x(x - 4) > 0$$ 8. **Analyze the inequality:** The product $x(x-4)$ is positive when both factors are positive or both are negative. - Case 1: $x > 0$ and $x - 4 > 0 \implies x > 4$ - Case 2: $x < 0$ and $x - 4 < 0 \implies x < 0$ 9. **Consider domain restrictions:** The original logarithms require $$x - 1 > 0 \implies x > 1$$ $$x - 3 > 0 \implies x > 3$$ So the domain is $x > 3$. 10. **Combine domain and solution:** From the domain $x > 3$ and the solution intervals, only $x > 4$ satisfies both. **Final answer:** $$\boxed{x > 4}$$