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    can anyone help me solve this equation:

    e^x + e^-x = 2

    Cheers
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    (Original post by spikey_aman)
    can anyone help me solve this equation:

    e^x + e^-x = 2

    Cheers
    surely just be inspection you can tell x=0 e^0+e^0=1+1=2
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    k if i'm correct, then log a + log b = log ab
    thefore, e^x + e^(-x) = log (-x^2)
    ..
    log (-x^2) = 2

    then just solve for x. k not sure if thats correct...soz..

    i guess is already been solved but what i've done above would that be the algebraic way 2 slove it??
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    (Original post by spikey_aman)
    can anyone help me solve this equation:

    e^x + e^-x = 2

    Cheers
    e^x+1/e^x=2

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

    e^2x-2e^x+1=0

    sub in y=e^x

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

    e^x=1
    x=0

    There may well be a quicker way, but I don't think I know it. You could also cheat by using hyperbolic functions:

    (e^x+e^-x)/2=cosh(x)
    2cosh(x)=2
    cosh(x)=1
    x=0
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    If you're familiar with hyperbolic functions, you're looking to solve cosh(x) = 1. Therefore:
    x = +/- arcosh(1) = 0
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    (Original post by dvs)
    If you're familiar with hyperbolic functions, you're looking to solve cosh(x) = 1. Therefore:
    x = +/- arcosh(1) = 0
    I'm sure that's cheating, though
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    (Original post by BovineBeast)
    I'm sure that's cheating, though
    :p:
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    Thanks BovineBeast - your method was the most helpful!
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    Yea, thats cheating - havent met the hyperbolics yet :P lol
 
 
 
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Updated: April 9, 2006
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