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Help with the number e

  • 05-05-2008 11:56AM
    #1
    IrishKnight
    Registered Users, Registered Users 2 Posts: 922 ✭✭✭


    To solve for x

    (e^x+e^-x)/2=1

    Any help on how to go about this?


Comments

  • #2
    Fremen
    Registered Users, Registered Users 2 Posts: 2,481 ✭✭✭Fremen


    Multiply both sides by two

    Multiply both sides by e^x

    Set y = e^x

    Solve the quadratic

    Use X = ln(y) to get your answer


  • #3
    IrishKnight
    Registered Users, Registered Users 2 Posts: 922 ✭✭✭IrishKnight


    Which would get me the following,

    e^x(e^x+e^-x)=2e^x
    (e^x)^2+1=2e^2

    Let y=e^2

    y^2-2y+1=0
    (y-1)(y-1)=0
    y=1

    e^x=1
    loge(1)=x
    x=0

    Which when I plug it back into the starting eqn I get 1, sweet. Thanks for that.


  • #4
    Michael Collins
    Moderators, Science, Health & Environment Moderators Posts: 1,852 Mod ✭✭✭✭Michael Collins


    Which, as you probably know already, is just arccosh(1) since

    (e^x+e^-x)/2 = cosh(x)


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