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

1 Answer
A. S. Adikesavan · EZ as pi
Jul 26, 2016

#x=log(5+-sqrt 21)/log 2 -1 = +-2.2604.#, nearly.

Explanation:

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

So, #y = 2^x = (5+-sqrt21)/2.# Both are positive and , therefore,

admissible. Equating logarithms and rearranging,

#x=log(5+-sqrt 21)/log 2 -1=+-2.2604#, nearly

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

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