2^x + 2^-x = 5/4 ... what is x?

Question

2^x + 2^-x = 5/4 ... what is x?

Thanks...

Answer 1

I believe the answer should be

log{ [ 5 +/- sqrt(39)i ]/8} / log 2.

Steps are as follows:

2^x + 1/2^x = 5/4
(2^x * 2^x + 1)/2^x = 5/4
4(2^x * 2^x + 1) = 5 * 2^x
4 * 2^x * 2^x + 4 = 5 * 2^x
4(2^x)^2 - 5(2^x) + 4 = 0

let 2^x be y

4y^2 - 5y + 4 = 0

solve it by the equation, you get

y = [5 +/- sqrt(25 - 4*4*4) ]/2*4
y = [5 +/- sqrt(-39) ]/8

since y = 2^x

2^x = [5 +/- sqrt(-39) ]/8
xlog(2) = log {[5 +/- sqrt(-39) ]/8}
x = log {[5 +/- sqrt(-39) ]/8} / log (2)
or
x = log {[5 +/- sqrt(39)i ]/8} / log (2)

Answer 2

2^x + 2^(-x) = (5/4)
2^x + (1/(2^x)) = (5/4)

Multiply everything by 4(2^x)

4(2^x)(2^x) + 1 = 5(2^x)
4(2^x)^2 - 5(2^x) + 4 = 0

this can be thought of as 4x^2 - 5x + 4 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-5) ± sqrt((-5)^2 - 4(4)(4)))/(2(4))
x = (5 ± sqrt(25 - 64))/8
x = (5 ± sqrt(-39))/8
x = (5 ± isqrt(39))/8

Now just plug 2^x back in for x

2^x = (5 ± isqrt(39))/8
x(ln(2)) = ln(5 ± isqrt(39))/8
x = (ln(5 ± isqrt(39))/8)/(ln(2))

ANS : (ln(5 ± isqrt(39))/(8ln2)

so there is no real solution.

If you go to www.quickmath.com, it will give you the same answer, but using log instead of ln