Question

Solve the following equations by reducing them to a quadratic equations:​

Answer 1

Answer:

X = 1

Step-by-step explanation:

${3}^{4x + 1} - 2 \times {3}^{2x + 2} - 81 = 0 \\ \\ let \: \: {3}^{2x} = y \\ \\ then \: \: \: \: 3 {y}^{2} - 2y(9) - 81 = 0 \\ \\ 3 {y}^{2} - 18y - 81 = 0 \\ \\ 3( {y}^{2} - 6y - 27) = 0 \\ \\ ( {y}^{2} - 6y - 27) = 0 \\ \\ {y}^{2} - 9y + 3y + 27 = 0 \\ \\ y(y - 9) + 3(y - 9) = 0 \\ \\ (y + 3)(y - 9) = 0 \\ \\ y = - 3 \: \: \: \: \: or \: \: \: \: \: \: \: y = 9 \\ \\ when \: \: y = - 3 \\ \\ {3}^{2x} = - 3 \: \: not \: possible \\ \\ and \: \: when \: \: \: y = 9 \\ \\ {3}^{2x} = 9 \\ \\ 2x = 2 \\ \\ x = 1.$