1. **State the problem:** Solve the exponential equation $$9^{3x+4} = 27^{4x+3}$$ for $x$. 2. **Recall the formula and rules:** Both 9 and 27 can be expressed as powers of 3 since $9 = 3^2$ and $27 = 3^3$. Using the property $a^{m} = a^{n} \implies m = n$ when $a > 0$ and $a \neq 1$, we can rewrite the equation with the same base and then equate the exponents. 3. **Rewrite the bases:** $$9^{3x+4} = (3^2)^{3x+4} = 3^{2(3x+4)} = 3^{6x + 8}$$ $$27^{4x+3} = (3^3)^{4x+3} = 3^{3(4x+3)} = 3^{12x + 9}$$ 4. **Set the exponents equal:** Since the bases are the same and positive, we have: $$6x + 8 = 12x + 9$$ 5. **Solve for $x$:** $$6x + 8 = 12x + 9$$ $$8 - 9 = 12x - 6x$$ $$-1 = 6x$$ $$x = -\frac{1}{6}$$ **Final answer:** $$x = -\frac{1}{6}$$