Question

2^2x+3 -9*2^x+1=0 find X

Answer 1

Answer:

${2}^{2x + 3} - 9 \times {2}^{x} + 1 = 0 \\ {2}^{2x} \times {2}^{3} - 9 \times {2}^{x} + 1 = 0 \\ 8 \times ({2}^{x} )^{2} - 9 \times {2}^{x} + 1 = 0 .....(1) \: \\ let \: {2}^{x} = y \\ \therefore \: equation \: (1) \: reduces \: to: \\ 8 {y}^{2} - 9y + 1 = 0 \\ 8 {y}^{2} - 8y - y+ 1 = 0 \\ 8y(y - 1) - 1(y - 1) = 0 \\ (y - 1)(8y - 1) = 0 \\ y - 1 = 0 \: or \: 8y - 1 = 0 \\ y = 1 \: or \: y = \frac{1}{8} \\ \because \: at \: y = 1 \: it \: is \: not \: possible \: to \\ find \: the \: value \: of \: x \: as \: bases \: will \\ be \: different : \\ so \: at \: \: \: \: \: \: y = \frac{1}{8} \\ {2}^{x} = \frac{1}{8} \\ {2}^{x} = \frac{1}{ {2}^{3} } \\ {2}^{x} = {2}^{ - 3} \\ x = - 3$

Answer 2

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