(9) \( 9^{x+1}-10.3^{x}+1=0 \)

To solve the equation: \[ 9^{x+1} - 10 \cdot 3^{x} + 1 = 0 \] **Step 1: Express \(9^{x+1}\) in terms of \(3^x\)** Recall that \(9 = 3^2\). Therefore, \[ 9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x + 2} = 3^2 \cdot 3^{2x} = 9 \cdot 3^{2x} \] Substituting back into the original equation: \[ 9 \cdot 3^{2x} - 10 \cdot 3^x + 1 = 0 \] **Step 2: Let \(u = 3^x\)** This substitution simplifies the equation. Let \(u = 3^x\). Then, \(3^{2x} = (3^x)^2 = u^2\). Substituting: \[ 9u^2 - 10u + 1 = 0 \] **Step 3: Solve the quadratic equation for \(u\)** The quadratic equation is: \[ 9u^2 - 10u + 1 = 0 \] Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9\), \(b = -10\), and \(c = 1\): \[ u = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} = \frac{10 \pm \sqrt{100 - 36}}{18} = \frac{10 \pm 8}{18} \] This gives two solutions: 1. \(u = \frac{10 + 8}{18} = \frac{18}{18} = 1\) 2. \(u = \frac{10 - 8}{18} = \frac{2}{18} = \frac{1}{9}\) **Step 4: Solve for \(x\)** Recall that \(u = 3^x\), so: 1. If \(u = 1\): \[ 3^x = 1 \implies x = 0 \] 2. If \(u = \frac{1}{9}\): \[ 3^x = \frac{1}{9} = 3^{-2} \implies x = -2 \] **Final Solutions:** \[ x = 0 \quad \text{and} \quad x = -2 \]