\( 5.2 ^ { x } + 3.2 ^ { x + 2 } = 68 \)

To solve the equation \( 5.2^{x} + 3.2^{x + 2} = 68 \), we can start by rewriting \( 3.2^{x + 2} \) as \( 3 \cdot 2^2 \cdot 2^{x} = 12 \cdot 2^{x} \). This gives us: \[ 5.2^{x} + 12 \cdot 2^{x} = 68 \] Now, let’s express \( 2^{x} \) as \( y \) (so \( 2^{x} = y \)), and rewrite the equation: \[ 5y + 12y^2 = 68 \] Next, move all terms to one side: \[ 12y^2 + 5y - 68 = 0 \] Now, we will use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 12 \), \( b = 5 \), and \( c = -68 \). Calculating the discriminant: \[ b^2 - 4ac = 5^2 - 4 \cdot 12 \cdot (-68) = 25 + 3264 = 3289 \] Now, compute \( y \): \[ y = \frac{-5 \pm \sqrt{3289}}{24} \] Approximating \( \sqrt{3289} \) gives about 57.34, so: \[ y \approx \frac{-5 \pm 57.34}{24} \] Calculating the positive root: \[ y \approx \frac{52.34}{24} \approx 2.18 \] Thus, we have \( 2^x \approx 2.18 \), leading to: \[ x \approx \log_2(2.18) \] Evaluating gives approximately: \[ x \approx 1.13 \] That’s our solution! \( x \approx 1.13 \). Happy calculating!