Question

please solve the sum​

Answer 1

Hey mate,

Here's your answer

${2}^{2x + 3} - 9 \times {2}^{x} + 1 = 0 \\ {2}^{2x} \times {2}^{3} - 9 \times {2}^{x} + 1 = 0$

Taking

${2}^{x}$

as x we get,

${x}^{2} \times 8 - 9 \times x + 1 = 0 \\ 8 {x}^{2} - 9x + 1 = 0 \\ {8x}^{2} - (8x + x) + 1 = 0 \\ {8x}^{2} - 8x + x + 1 = 0 \\ 8x(x - 1) - 1(x - 1) = 0\\ (x - 1)(8x - 1) = 0 \\ either \\ x - 1 = 0 \\ x = 1 \\ or \\ 8x - 1 = 0 \\ 8x = 1 \\ x = 1 \div 8$

Hope it helps ,

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