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Turning arcsin into logs

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Hello. Will someone please explain to me why this is true?
    arcsinh(e^x) = ln(e^x + √(e^(2x) + 1))



    2. The attempt at a solution

    I cannot figure out why arcsin is able to be put into terms of ln. Thank you in advance.
     
  2. jcsd
  3. Feb 24, 2014 #2

    tiny-tim

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    hi caty! :smile:

    (hey, what's an h ? :wink:)

    2sinh[ln(ex + √(e2x + 1))]

    = exp[ln(ex + √(e2x + 1))] - exp[-ln(ex - √(e2x + 1))]

    = ex + √(e2x + 1)) - 1/[ex + √(e2x + 1))]

    = [e2x + e2x + 1 + 2ex√(e2x + 1)) - 1]/[ex + √(e2x + 1))]

    = 2ex :wink:

    alternatively, put ex = sinhy

    then sinh[ln(ex + √(e2x + 1))]

    = sinh[ln(sinhy + coshy)]

    = sinh[ln(ey)]

    = sinh[y] = ex
     
  4. Feb 24, 2014 #3

    LCKurtz

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    In addition to Tiny's explanation you could also note ##y =\sinh^{-1}(e^x)## is the same as ##e^x =\sinh(y)=
    \frac{e^y - e^{-y}} 2## or ##2e^x = e^y - \frac 1 {e^y}##. Solve that for ##y## in terms of ##x## and you will get your expression and you will see where the logarithms come from.
     
    Last edited: Feb 24, 2014
  5. Feb 25, 2014 #4
    Ahh got it! Thank you very much!
     
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