How do you solve $2^{x} + 2^{- x} = 5$ $2^x+2^(-x)=5$ ?

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Answer 1 · 2016-07-26T13:28:14

$x = \frac{log (5 \pm \sqrt{21})}{log 2} - 1 = \pm 2.2604 .$ $x=log(5+-sqrt 21)/log 2 -1 = +-2.2604.$ , nearly.

This is a quadratic equation $y^{2} - 5 y + 1 = 0$ $y^2-5y+1=0$ , where $y = 2^{x}$ $y = 2^x$ .

So, $y = 2^{x} = \frac{5 \pm \sqrt{21}}{2} .$ $y = 2^x = (5+-sqrt21)/2.$ Both are positive and , therefore,

admissible. Equating logarithms and rearranging,

$x = \frac{log (5 \pm \sqrt{21})}{log 2} - 1 = \pm 2.2604$ $x=log(5+-sqrt 21)/log 2 -1=+-2.2604$ , nearly

The given equation has an alternative form cosh(x log 2) = 2.5.

How do you solve 2x+2−x=5 2^x+2^(-x)=5?