b) \( 2^{x}+\frac{32}{2^{x}}=12 \)

To solve the equation \( 2^{x}+\frac{32}{2^{x}}=12 \), let's start by making a substitution. Let \( y = 2^{x} \). Then, the equation becomes: \[ y + \frac{32}{y} = 12 \] Multiplying the entire equation by \( y \) to eliminate the fraction gives: \[ y^2 + 32 = 12y \] Rearranging the equation leads to the standard quadratic form: \[ y^2 - 12y + 32 = 0 \] Now, we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -12, c = 32 \): \[ b^2 - 4ac = (-12)^2 - 4 \cdot 1 \cdot 32 = 144 - 128 = 16 \] Now, applying the values in the quadratic formula: \[ y = \frac{12 \pm \sqrt{16}}{2 \cdot 1} = \frac{12 \pm 4}{2} \] Calculating the two potential solutions: \[ y = \frac{16}{2} = 8 \quad \text{and} \quad y = \frac{8}{2} = 4 \] Since \( y = 2^{x} \), we substitute back to find \( x \): 1. For \( y = 8 \): \[ 2^{x} = 8 \Rightarrow 2^{x} = 2^3 \Rightarrow x = 3 \] 2. For \( y = 4 \): \[ 2^{x} = 4 \Rightarrow 2^{x} = 2^2 \Rightarrow x = 2 \] So the solutions to the original equation are \( x = 3 \) and \( x = 2 \).