1. **State the problem:** Solve for $x$ in the equation $$\log_3(3x - 2) = 2.$$\n\n2. **Recall the logarithm definition:** The equation $\log_a b = c$ means $$a^c = b.$$\n\n3. **Apply the definition:** Here, $a=3$, $b=3x - 2$, and $c=2$, so $$3^2 = 3x - 2.$$\n\n4. **Calculate the power:** $$9 = 3x - 2.$$\n\n5. **Solve for $x$:** Add 2 to both sides: $$9 + 2 = 3x,$$ which simplifies to $$11 = 3x.$$\n\n6. **Divide both sides by 3:** $$\frac{\cancel{11}}{\cancel{3}} = x,$$ so $$x = \frac{11}{3}.$$\n\n7. **Check the domain:** The argument of the logarithm must be positive: $$3x - 2 > 0 \Rightarrow 3x > 2 \Rightarrow x > \frac{2}{3}.$$\nSince $\frac{11}{3} > \frac{2}{3}$, the solution is valid.\n\n**Final answer:** $$x = \frac{11}{3}.$$