HOW DO YOU DO THIS math problem 2^x+2^-x=5?

Question

2^x+2^-x=5 ????? solve for x

Answer 1

2^x + 2^-x = 5

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

Therefore

2^x = a
1/(2^x) = 1/a
a + 1/a = 5

a^2 + 1 = 5a
a^2 - 5a + 1 = 0

a = [5 ± √(25 - 4)]/2 Quadratic formula
a = 5 ± √(21)/2
a ≈ 4.79 or 0.21

2^x = a
x = log 2(a)

x = log 2(4.79) or x = log 2(0.21)

x ≈ ±2.26

Answer 2

2 times 5 plus 2 times 5= 20

10 plus 10= 20

answer= 20.

Answer 3

X b 2^x then eq become
2^2x - 5*2^x + 1 = 0
let y = 2^x
y^2 - 5y + 1 =0

use the quad formula to solve for y, then eval x from there

Answer 4

between the properties of multiplication is that, if a made from words = 0, a minimum of between the words could desire to = 0. once you have factored your equation into 2x(x+5)=0, this potential that x=0 or (x+5) = 0, which in turn potential that x= -5

Answer 5

2^x+2^-x=5
[2^(2x) +1]/2^x=5
2^(2x) +1 = 5*2^x
2^2x - 5*2^x + 1 = 0
let y = 2^x
Then y^2 -5y +1 = 0
y = [5 +/-sqrt(25-4}/2
y = 2.5 +/- .5sqrt(21
So 2^x = 2.5 +/- .5sqrt(21)
xln2 = ln{2.5 +/- .5sqrt(21)}
x = [ln{2.5 +/- .5sqrt(21)}]/ln2
x approximately = +/- 2.26

Answer 6

simple...

2 + 2 = 4
then move on to the 5
5 x 4 = 20

ANSWER.20

Answer 7

hard