Question

2^(x+3)+2^(-x)=6
then what is the value of x?

Answer 1

${2}^{x + 3} + {2}^{ - x} = 6 \\ {2}^{x} \times {2}^{3} + \frac{1}{ {2}^{x} } = 6 \\ 8 \times {2}^{x} + \frac{1}{ {2}^{x} } = 6 \\ let \: {2}^{x} = y \\ 8y + \frac{1}{y} = 6 \\ 8 {y}^{2} - 6y + 1 = 0 \\ 8 {y}^{2} - 4y - 2y + 1 = 0 \\ 4y(2y - 1) - (2y - 1) = 0 \\ (4y - 1)(2y - 1) = 0 \\ y = \frac{1}{4} \: and \: \frac{1}{2} \\ now \: \\ {2}^{x} = \frac{1}{4} \\ {2}^{x} = {2}^{ - 2} \\ \green{\bold{x = - 2}} \\ {2}^{x} = \frac{1}{2} \\ {2}^{x} = {2}^{ - 1} \\ \bold{\red{x = - 1}}$

so answer is -2,-1

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