1. **State the problem:** Solve the logarithmic equation $$\log_4(3x + 10) = 2$$. 2. **Recall the definition of logarithm:** If $$\log_b(a) = c$$, then $$a = b^c$$. 3. **Apply the definition:** From $$\log_4(3x + 10) = 2$$, we get $$3x + 10 = 4^2$$. 4. **Calculate the power:** $$4^2 = 16$$, so $$3x + 10 = 16$$. 5. **Solve for $$x$$:** $$3x = 16 - 10$$ $$3x = 6$$ 6. **Divide both sides by 3:** $$x = \frac{6}{3}$$ 7. **Show cancellation:** $$x = \frac{\cancel{6}}{\cancel{3}} = 2$$. 8. **Check the domain:** The argument of the logarithm must be positive: $$3x + 10 > 0 \implies 3(2) + 10 = 6 + 10 = 16 > 0$$, which is true. 9. **Final answer:** $$x = 2$$.