Logarithmic form of hyperbolic functions

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  1. music lover's Avatar
    1 01-08-2012  15:47
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    Logarithmic form of hyperbolic functions
    Bit stuck on following the derivation: y=coshx;x=coshy
    coshy=\frac{1}{2}(e^y+e^-^y);x=\frac{1}{2}(e^y+e^-^y). So then we get: e^y-2x+e^-^y=0
    e^2^y-2xe^y+1=0. This next step I don't understand. How do we get to: e^y=x\pm\sqrt(x^2-1)
  2. ian.slater's Avatar
    2 01-08-2012  15:49
    • Exalted and Worshipped Member
    Re: Logarithmic form of hyperbolic functions
    (Original post by music lover)
    Bit stuck on following the derivation: y=coshx;x=coshy
    coshy=\frac{1}{2}(e^y+e^-^y);x=\frac{1}{2}(e^y+e^-^y). So then we get: e^y-2x+e^-^y=0
    e^2^y-2xe^y+1=0. This next step I don't understand. How do we get to: e^y=x\pm\sqrt(x^2-1)
    Solve the quadratic in e^y using the usual formula
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  3. music lover's Avatar
    3 01-08-2012  16:33
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    • Posts: 146
    Re: Logarithmic form of hyperbolic functions
    (Original post by ian.slater)
    Solve the quadratic in e^y using the usual formula
    Oh yeah, thanks
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