h) \( 4^{x}-2^{x+1}=48 \)

To tackle the equation \( 4^{x}-2^{x+1}=48 \), we can start by expressing \( 4^{x} \) in terms of \( 2^{x} \). Since \( 4^{x} = (2^{2})^{x} = 2^{2x} \), we can rewrite the equation as: \[ 2^{2x} - 2 \cdot 2^x = 48 \] Now, let \( y = 2^{x} \). This means \( 2^{2x} = y^2 \), so we have: \[ y^2 - 2y - 48 = 0 \] Now, we can factor this quadratic equation: \[ (y - 8)(y + 6) = 0 \] Setting each factor to zero gives us: \[ y - 8 = 0 \quad \Rightarrow \quad y = 8 \] \[ y + 6 = 0 \quad \Rightarrow \quad y = -6 \quad (\text{discarded since } y = 2^{x} \text{ cannot be negative}) \] Now we know \( y = 8 \), which gives us: \[ 2^{x} = 8 \quad \Rightarrow \quad 2^{x} = 2^{3} \quad \Rightarrow \quad x = 3 \] So, the solution to your equation is \( x = 3 \).