Find the value of \( x \) in \( 9^{x}-4 \times 3^{x+1}+27=0 \) tt: take \( P=3^{x} \)

To solve the equation: \[ 9^{x} - 4 \times 3^{x+1} + 27 = 0 \] Let's follow the suggested substitution \( P = 3^{x} \). 1. **Express all terms in terms of \( P \):** - \( 9^{x} = (3^2)^{x} = 3^{2x} = (3^{x})^2 = P^2 \) - \( 3^{x+1} = 3^{x} \times 3 = 3P \) Substituting these into the original equation: \[ P^2 - 4 \times 3P + 27 = 0 \\ P^2 - 12P + 27 = 0 \] 2. **Solve the quadratic equation for \( P \):** \[ P^2 - 12P + 27 = 0 \] Using the quadratic formula \( P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): - \( a = 1 \) - \( b = -12 \) - \( c = 27 \) \[ P = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 1 \times 27}}{2 \times 1} \\ P = \frac{12 \pm \sqrt{144 - 108}}{2} \\ P = \frac{12 \pm \sqrt{36}}{2} \\ P = \frac{12 \pm 6}{2} \] This gives two solutions for \( P \): \[ P = \frac{18}{2} = 9 \quad \text{or} \quad P = \frac{6}{2} = 3 \] 3. **Solve for \( x \) using \( P = 3^x \):** - **First Solution:** \[ 3^x = 9 \\ 3^x = 3^2 \\ x = 2 \] - **Second Solution:** \[ 3^x = 3 \\ 3^x = 3^1 \\ x = 1 \] **Final Solutions:** \[ x = 1 \quad \text{and} \quad x = 2 \] Answer: Both solutions are \( x = 1 \) and \( x = 2 \).