Solve for \( x \) : \( 3^{2 x+2}-10.3^{x}+1=0 \)

To solve the equation \( 3^{2x+2} - 10 \cdot 3^x + 1 = 0 \), we can start by making a substitution. Let \( y = 3^x \). Then, we can rewrite \( 3^{2x} \) as \( y^2 \) because \( 3^{2x} = (3^x)^2 = y^2 \). Thus, the equation transforms into: \[ y^2 \cdot 3^2 - 10y + 1 = 0 \] This simplifies to: \[ 9y^2 - 10y + 1 = 0 \] Now, we can solve this quadratic equation using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 9 \), \( b = -10 \), and \( c = 1 \). Calculating the discriminant: \[ b^2 - 4ac = (-10)^2 - 4 \cdot 9 \cdot 1 = 100 - 36 = 64 \] Since the discriminant is positive, we have two distinct real solutions for \( y \): \[ y = \frac{10 \pm \sqrt{64}}{18} = \frac{10 \pm 8}{18} \] This gives us two values: 1. \( y = \frac{18}{18} = 1 \) 2. \( y = \frac{2}{18} = \frac{1}{9} \) Now we rewrite these back in terms of \( 3^x \): 1. For \( 3^x = 1 \), we have \( x = 0 \). 2. For \( 3^x = \frac{1}{9} \), we know \( \frac{1}{9} = 3^{-2} \) which gives us \( x = -2 \). Thus, the solutions are: \[ x = 0 \quad \text{and} \quad x = -2 \]