1. **State the problem:** Solve the logarithmic equation $$2 \log_3 x - \log_3 (x - 2) = 2$$ for $x$. 2. **Recall logarithm properties:** - $a \log_b c = \log_b c^a$ - $\log_b A - \log_b B = \log_b \frac{A}{B}$ 3. **Apply the power rule:** $$2 \log_3 x = \log_3 x^2$$ 4. **Rewrite the equation using the subtraction rule:** $$\log_3 x^2 - \log_3 (x - 2) = \log_3 \frac{x^2}{x - 2}$$ 5. **Set the equation equal to 2:** $$\log_3 \frac{x^2}{x - 2} = 2$$ 6. **Convert logarithmic form to exponential form:** $$\frac{x^2}{x - 2} = 3^2$$ $$\frac{x^2}{x - 2} = 9$$ 7. **Solve the rational equation:** $$x^2 = 9(x - 2)$$ $$x^2 = 9x - 18$$ 8. **Bring all terms to one side:** $$x^2 - 9x + 18 = 0$$ 9. **Factor the quadratic:** $$ (x - 6)(x - 3) = 0$$ 10. **Find the roots:** $$x = 6 \quad \text{or} \quad x = 3$$ 11. **Check domain restrictions:** - $x > 0$ because $\log_3 x$ is defined. - $x - 2 > 0 \Rightarrow x > 2$. 12. **Check solutions:** - $x = 3$ satisfies $x > 2$. - $x = 6$ satisfies $x > 2$. 13. **Final answer:** $$\boxed{x = 3 \text{ or } x = 6}$$