1. **State the problem:** Solve the system of equations: $$9(9^x) = 27^{y+2}$$ $$4^x \times 8^y = i$$ 2. **Rewrite the bases as powers of primes:** - $9 = 3^2$ - $27 = 3^3$ - $4 = 2^2$ - $8 = 2^3$ 3. **Rewrite each term:** $$9(9^x) = 9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2}$$ $$27^{y+2} = (3^3)^{y+2} = 3^{3(y+2)} = 3^{3y+6}$$ 4. **Set the exponents equal since bases are the same:** $$2x + 2 = 3y + 6$$ 5. **Rewrite the second equation:** $$4^x \times 8^y = (2^2)^x \times (2^3)^y = 2^{2x} \times 2^{3y} = 2^{2x + 3y} = i$$ 6. **Express $i$ in exponential form:** $i$ is the imaginary unit, $i = e^{i\pi/2}$ in complex exponential form, but since the right side is a complex number and the left side is a positive real number (powers of 2), this equation has no solution in real numbers. 7. **Summary:** - From the first equation: $$2x + 2 = 3y + 6 \Rightarrow 2x - 3y = 4$$ - The second equation implies $2^{2x + 3y} = i$, which is impossible for real $x,y$. **Final answer:** No real solutions exist for $x$ and $y$ satisfying both equations simultaneously. If complex solutions are considered, further analysis with complex logarithms is needed.