solve for x: 2^(x) + 2^(-x) = 5?

Answer 1

Let y = 2^x
y + 1 / y = 5
y² + 1 = 5y
y² - 5y + 1 = 0
y = 2^x = [ 5 ± √21 ] / 2
x ln 2 = [ 5 ± √21] / 2
x = [ (1 / (2 ln 2) ] [5 ± √21 ]

Answer 2

2^(x) + 2^(-x) = 5

Multiplying both sides by 2^x we get

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

Let 2^x = u, then 2^(2x) = u^2

Substituting into (1) we get
u^2 - 5u + 1 = 0

(u - 5/2)^2 - 25/4 + 1 = 0
(u - 5/2)^2 = 21/4

u - 5/2 = +/- sqrt(21)/2
u = [5 +/- sqrt(21)]/2

When u = [5 + sqrt(21)]/2
u = 4.791

2^x = 4.791

Taking ln() of both sides

x ln(2) = ln(4.791)
x = 2.26

Similarly when u = [5 - sqrt(21)]/2 you will get

x = -2.26

Therefore,
x = +/- 2.26

Answer 3

2^x[2^x + 2^(-x) -5] = 0
2^(2x) - 5(2^x) + 1 = 0

set y = 2^x

y^2 - 5y + 1 = 0
y = 5/2 +/- sqrt(25 - 4)/2 = (5 +/- sqrt(21))/2

take the logarithm and employ y = 2^x

xlog(2) = log(5 +/- sqrt(21)) - log(2)

x = log(5 +/- sqrt(21))/log(2) - 1, your two solutions.

Answer 4

2^(x) - 5 + 2^(-x) = 0
2^(-x)[2^(2x) - 5*2^(x) + 1] = 0
Use quadratic formula,
2^(x) = (5±√21)/2
x = ln[ (5±√21)/2]/ln2 = ±2.260

Answer 5

(x)^2 + (-x)^2 = 5
x^2 + x^2 = 5
2x^2 = 5
x^2 = 5/2
x = +-sqrt(5/2)