I know you need to use logarithms to solve the equation, but i get lost or am missing a step if anyone can walk me through it, not necessarily solve, that would be great!
2^x + 2^-x = 8
2^x + 2^-x = 8
-
2^x + 2^(-x) = 8
2^x + 1/(2^x) = 8
Here's where it gets fun. Let 2^x = t
t + 1/t = 8
Multiply everything by t
t^2 + 1 = 8t
t^2 - 8t + 1 = 0
t = (8 +/- sqrt(64 - 4)) / 2
t = (8 +/- 2 * sqrt(15)) / 2
t = 4 +/- sqrt(15)
2^x = 4 +/- sqrt(15)
x * ln(2) = ln(4 + sqrt(15)) , ln(4 - sqrt(15))
x = ln(4 + sqrt(15)) / ln(2) , ln(4 - sqrt(15)) / ln(2)
2^x + 1/(2^x) = 8
Here's where it gets fun. Let 2^x = t
t + 1/t = 8
Multiply everything by t
t^2 + 1 = 8t
t^2 - 8t + 1 = 0
t = (8 +/- sqrt(64 - 4)) / 2
t = (8 +/- 2 * sqrt(15)) / 2
t = 4 +/- sqrt(15)
2^x = 4 +/- sqrt(15)
x * ln(2) = ln(4 + sqrt(15)) , ln(4 - sqrt(15))
x = ln(4 + sqrt(15)) / ln(2) , ln(4 - sqrt(15)) / ln(2)