Answer:
No solutions
Step-by-step explanation:
[tex]9^x + 4(3^x + 1) + 27 = 0\\3^{2x} + 4 \cdot 3^x + 31 = 0[/tex]
Substitute 3^x with t and we get.
[tex]t^2 + 4t + 31 = 0\\\triangle = 16 - 4 \cdot 31 = -108 < 0 => t_{12} \notin R => x \notin R[/tex]
