Solve for x: e^x+e^-x=3?

Question

I have the answer just need to know how to solve the question.

answer is x=ln((3+ or - radical 5)/2).

Answer 1

if you know what hyperbolic functino are you cna substitute cosh(x) for e^x+e^-x to get cosh(x)=3 then x = cosh^-1(3) and cosh^-1=ln(x+sqrt(x^2-1)) for x>=1

by the way cosh(x)=cos( ix) where i= sqrt(-1)

Answer 2

Let y = e^x:-

y + 1/y = 3

y² + 1 = 3y

y² - 3y + 1 = 0

y = [ 3 ± √ (9 - 4) ] / 2

y = [ 3 ± √5 ] / 2

e^x = [ 3 ± √5 ] / 2

x = ln [ 3 ± √5 ] / 2  

Answer 3

This is a quadratic equation in disguise. Let y = e^x. Then, we have

y + 1/y = 3

Multiply both sides by y, giving

y^2 + 1 = 3y

Therefore,

y^2 - 3y + 1 = 0

using the quadratic formula gives,

y = [3±sqrt(5)]/2

and

x = ln y.