Question 7 ind the solution set of \[ 3^{2(x+1)}-8.3^{x+1}=9 \]

To solve the equation \( 3^{2(x+1)} - 8 \cdot 3^{x+1} = 9 \), we can start by substituting \( y = 3^{x+1} \). This gives us: \[ 3^{2(x+1)} = (3^{x+1})^2 = y^2 \] Thus, the equation becomes: \[ y^2 - 8y - 9 = 0 \] Next, we can factor this quadratic equation or use the quadratic formula: \[ y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} \] \[ y = \frac{8 \pm \sqrt{64 + 36}}{2} \] \[ y = \frac{8 \pm \sqrt{100}}{2} \] \[ y = \frac{8 \pm 10}{2} \] This gives us two potential solutions for \( y \): \[ y = \frac{18}{2} = 9 \quad \text{and} \quad y = \frac{-2}{2} = -1 \] Since \( y = 3^{x+1} \) must be positive, we discard \( y = -1 \). Thus, we have: \[ 3^{x+1} = 9 \] This can be rewritten as: \[ 3^{x+1} = 3^2 \] So, equating the exponents: \[ x + 1 = 2 \quad \Rightarrow \quad x = 1 \] Thus, the solution set is \( \{1\} \). If you want to explore more about the exponential equations or their applications in real life, just let me know!