1. **State the problem:** Solve the inequality $x \ln(x-1) - 2 > 0$. 2. **Rewrite the inequality:** $$x \ln(x-1) > 2$$ 3. **Domain considerations:** Since $\ln(x-1)$ is defined only for $x-1 > 0$, we have: $$x > 1$$ 4. **Analyze the function:** Define $f(x) = x \ln(x-1)$. We want to find where $f(x) > 2$ for $x > 1$. 5. **Check behavior near domain start:** As $x \to 1^+$, $\ln(x-1) \to -\infty$, so $f(x) \to 1 \cdot (-\infty) = -\infty$. 6. **Check for critical points:** Find $f'(x)$: $$f'(x) = \ln(x-1) + \frac{x}{x-1}$$ 7. **Set derivative to zero to find extrema:** $$\ln(x-1) + \frac{x}{x-1} = 0$$ 8. **Solve for $x$ numerically:** This transcendental equation has a unique minimum at some $x > 1$. Approximate numerically (e.g., $x \approx 1.8$). 9. **Check values of $f(x)$ at critical points and at large $x$:** - At $x=2$, $f(2) = 2 \ln(1) = 0$. - At $x=3$, $f(3) = 3 \ln(2) \approx 3 \times 0.693 = 2.079 > 2$. 10. **Find exact solution for $f(x) = 2$:** Solve: $$x \ln(x-1) = 2$$ 11. **Use substitution:** Let $t = x-1$, so $x = t+1$, $t > 0$. Equation becomes: $$(t+1) \ln t = 2$$ 12. **Numerical approximation:** Try $t=1.5$: $$(1.5+1) \ln 1.5 = 2.5 \times 0.405 = 1.0125 < 2$$ Try $t=3$: $$(3+1) \ln 3 = 4 \times 1.0986 = 4.3944 > 2$$ Try $t=1.8$: $$(1.8+1) \ln 1.8 = 2.8 \times 0.5878 = 1.646 < 2$$ Try $t=2.5$: $$(2.5+1) \ln 2.5 = 3.5 \times 0.9163 = 3.207 > 2$$ Try $t=2$: $$(2+1) \ln 2 = 3 \times 0.693 = 2.079 > 2$$ Try $t=1.9$: $$(1.9+1) \ln 1.9 = 2.9 \times 0.6419 = 1.862 < 2$$ Try $t=1.95$: $$(1.95+1) \ln 1.95 = 2.95 \times 0.6678 = 1.969 < 2$$ Try $t=1.98$: $$(1.98+1) \ln 1.98 = 2.98 \times 0.6831 = 2.035 > 2$$ So root $t_0 \approx 1.97$. 13. **Final solution:** Since $f(x)$ is increasing for $x >$ minimum point and $f(x) > 2$ for $x > x_0 = t_0 + 1 \approx 2.97$, the solution to the inequality is: $$\boxed{x > 2.97}$$ **Summary:** - Domain: $x > 1$ - Inequality holds for $x > 2.97$ approximately.