1. **State the problem:** Solve for $n$ in the equation $$9^{2n-1} = 27^{n+2}.$$ 2. **Rewrite bases as powers of the same base:** Both 9 and 27 can be written as powers of 3: $$9 = 3^2, \quad 27 = 3^3.$$ So the equation becomes $$\left(3^2\right)^{2n-1} = \left(3^3\right)^{n+2}.$$ 3. **Use the power of a power rule:** $$\left(a^m\right)^n = a^{mn}.$$ Applying this, we get $$3^{2(2n-1)} = 3^{3(n+2)}.$$ 4. **Simplify exponents:** $$3^{4n - 2} = 3^{3n + 6}.$$ 5. **Since the bases are equal and nonzero, set exponents equal:** $$4n - 2 = 3n + 6.$$ 6. **Solve for $n$: $$4n - 2 = 3n + 6$$ $$4n - \cancel{2} - 3n = 3n + 6 - 3n$$ $$n - 2 = 6$$ $$n = 6 + 2$$ $$n = 8.$$ **Final answer:** $n = 8$ which corresponds to option D.