1. **State the problem:** Solve for $x$ in the equation $\log_3(2x-5) = 2$. 2. **Recall the logarithm definition:** $\log_b(a) = c$ means $b^c = a$. 3. **Apply the definition:** From $\log_3(2x-5) = 2$, we get $$3^2 = 2x - 5$$ 4. **Calculate the power:** $$9 = 2x - 5$$ 5. **Isolate $x$:** $$9 + 5 = 2x$$ $$14 = 2x$$ 6. **Divide both sides by 2:** $$\cancel{2}x = \frac{14}{\cancel{2}}$$ $$x = 7$$ 7. **Check the domain:** The argument of the logarithm must be positive: $$2x - 5 > 0$$ $$2(7) - 5 = 14 - 5 = 9 > 0$$ So $x=7$ is valid. **Final answer:** $$x = 7$$