Solve for $9^x-90 \times 3^x+729=0$

A. 1, 2

B. 2, 3

C. 3, 5

D. 2, 4

D. 2,4

$\begin{aligned}
& 9^x-90 \times 3^x+729=0 \\
\Rightarrow & \left(3^x\right)^2-81 \cdot 3^x-9 \cdot 3^x+729=0 \\
\Rightarrow & 3^x\left(3^x-81\right)-9\left(3^x-81\right)=0
\end{aligned}$

$3^x =81 or 9$

$x=2,4$