1. **State the problem:** Solve the inequality $\log x > \log (x-1)$. 2. **Recall the properties of logarithms:** The logarithm function $\log x$ is increasing for $x > 0$. This means if $\log a > \log b$, then $a > b$ provided $a > 0$ and $b > 0$. 3. **Apply the property:** Since $\log x > \log (x-1)$, it implies $$x > x - 1.$$ 4. **Simplify the inequality:** $$x > x - 1 \implies x - x > -1 \implies 0 > -1,$$ which is always true. 5. **Consider the domain restrictions:** For $\log x$ and $\log (x-1)$ to be defined, we need $$x > 0 \quad \text{and} \quad x - 1 > 0 \implies x > 1.$$ 6. **Combine the domain and inequality:** Since the inequality is always true, the solution is the domain where both logarithms are defined: $$\boxed{x > 1}.$$