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

To solve the equation \( 2^{2x+1} + 4^{x+1} = 2^{2x} \left( \frac{9}{2} \right) \), we can start by rewriting \( 4^{x+1} \) as \( (2^2)^{x+1} = 2^{2(x+1)} = 2^{2x+2} \). Substituting this into the equation gives us: \[ 2^{2x+1} + 2^{2x + 2} = 2^{2x} \cdot \frac{9}{2}. \] Now, we can factor out \( 2^{2x} \) from the left-hand side: \[ 2^{2x} \left( 2 + 4 \right) = 2^{2x} \cdot \frac{9}{2}. \] This simplifies to: \[ 2^{2x} \cdot 6 = 2^{2x} \cdot \frac{9}{2}. \] Assuming \( 2^{2x} \neq 0 \), we can divide both sides by \( 2^{2x} \): \[ 6 = \frac{9}{2}. \] However, this is not true. This suggests that our assumption might lead us to a case where there are no solutions. To explore further, let’s check if \( 2^{2x} \) could be equal to zero, which is impossible, as the exponential function never equals zero. Thus there are no values of \( x \) that can satisfy the original equation. So, the equation has no solution.