(C) \( 2^{2 x-1}+4^{x+1}=2^{2 x}\left(\frac{9}{2}\right) \)

To solve the equation \( 2^{2 x-1} + 4^{x+1} = 2^{2 x}\left(\frac{9}{2}\right) \), we can start by rewriting \( 4^{x+1} \) in terms of base 2. Since \( 4 = 2^2 \), we have: \[ 4^{x+1} = (2^2)^{x+1} = 2^{2(x + 1)} = 2^{2x + 2}. \] Thus, the equation becomes: \[ 2^{2x-1} + 2^{2x + 2} = 2^{2x}\left(\frac{9}{2}\right). \] Now, we can factor \( 2^{2x} \) from the left side: \[ 2^{2x - 1} + 2^{2x + 2} = 2^{2x}(2^{-1} + 2^{2}). \] Calculating \( 2^{-1} + 2^{2} \): \[ 2^{-1} = \frac{1}{2}, \quad 2^{2} = 4 \quad \Rightarrow \quad \frac{1}{2} + 4 = \frac{1}{2} + \frac{8}{2} = \frac{9}{2}. \] So the left side can be rewritten as: \[ 2^{2x} \left(\frac{9}{2}\right). \] Now the equation is: \[ 2^{2x}\left(\frac{9}{2}\right) = 2^{2x}\left(\frac{9}{2}\right). \] Since both sides are equal, the equation holds for any value of \( x \). Thus, the solution is: \[ \text{All real } x. \]