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Math Help - calc II, but stuck on algebra

  1. #1
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    Sep 2005
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    Question a problem for calc II, but stuck on algebra

    alright, so in this problem i need to solve for A.

    y = 4e^x + e^-x
    y = Acosh(x + ln2)

    now you might already know cosh(x) = (e^x + e^-x) / 2

    and with a little fooling around i found that e ^ ln2 = 2
    and that cosh(ln(2)) = 1.25 (?)

    so i think i have most of the pieces figured out but i cannot put them together. after the substitution where should i go with this?
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  2. #2
    MHF Contributor
    Joined
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    Let me use the datas/informations as posted. If they are wrongly posted, then my answer will be wrong also.

    y = 4e^x + e^-x -----------(1)
    y = A*cosh(x + ln2) --------------(2)

    -----------------
    cosh(ln(2))
    = [e^ln(2) +e^(-ln(2))]/2 ----(i)

    Let e^ln(2) = N
    Take the ln of both sides,
    ln(2)*ln(e) = ln(N)
    ln(2) = ln(N)
    2 = N
    So, e^ln(2) = 2 -----***

    Let e^[-ln(2)] = M
    1/[e^ln(2)] = M
    1/[2] = M
    So, e^[-ln(2)] = 1/2 -----***

    Substitute those into the cosh(ln(2)),
    cosh(ln(2))
    = [e^ln(2) +e^(-ln(2))]/2 ----(i)
    = [2 +1/2]/2
    = 5/4 or 1.25 ---------***

    -----------------------------
    sinh(ln(2))
    = [e^ln(2) -e^(-ln(2))]/2 ----(ii)
    = [2 -1/2]/2
    = 3/4 or 0.75 ---------***

    ---------------------------------------------
    y = A*cosh(x + ln2) --------------(2)
    y = A*{cosh(x)cosh[ln(2)] +sinh(x)sinh[ln(2)]}
    y = A*{[(e^x +e^-x)/2](5/4) +[(e^x -e^-x)/2](3/4)}
    y = (A/8){(e^x +e^-x)(5) +(e^x -e^-x)(3)}
    y = (A/8){5e^x +5e^-x +3e^x -3e^-x}
    y = (A/8){8e^x +2e^-x}
    y = (A/4){4e^x +e^-x} ----------(2a)

    y from (2a) = y from (1),
    (A/4){4e^x +e^-x} = 4e^x +e^-x
    Divide both sides by (4e^x +e^-x),
    A/4 = 1
    A = 1*4 = 4 --------------------------answer.
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