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Math Help - simplify logarithm

  1. #1
    jas
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    simplify logarithm

    hi
    i have the following
    ln t = B1 + B2.T + B3.X + B4.X^2 + B5.X^3
    where B1,B2,B3,B4,B5 are constants and X = ln y
    i need this in a form t = C1 (y^C2) (e^(-C3/T))
    where C1,C2,C3 are constants.
    ln and e are natural log terms.
    can this be done?
    thanks
    jas
    p.s. i know e^(B3.X) = e^(B3. ln y) =e^(ln y^B3) = y^B3
    but i'm not sure how to get the other terms B4.X^2 = B4 (ln y) (ln y)
    and B5.X^3 = B5 (ln y) (ln y) (ln y)
    in a similar form of y^??? (or if its possible!!)
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  2. #2
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    Quote Originally Posted by jas View Post
    hi
    i have the following
    ln t = B1 + B2.T + B3.X + B4.X^2 + B5.X^3
    where B1,B2,B3,B4,B5 are constants and X = ln y
    i need this in a form t = C1 (y^C2) (e^(-C3/T))
    where C1,C2,C3 are constants.
    ln and e are natural log terms.
    can this be done?
    thanks
    jas
    p.s. i know e^(B3.X) = e^(B3. ln y) =e^(ln y^B3) = y^B3
    but i'm not sure how to get the other terms B4.X^2 = B4 (ln y) (ln y)
    and B5.X^3 = B5 (ln y) (ln y) (ln y)
    in a similar form of y^??? (or if its possible!!)
    Taking the exponential of both sides,
    t= e^{B1+ B2T+ B3X+ B4X^2+ B5X^3}= e^{B1}(e^{T})^B2(e^x)^{B3}(e^X)^{2B4}(e^X)^{3B5}
    Since x= ln y, e^X= e^{ln y}= y so that is the same as t= e^{B1}e^{B2T}y^{B3}y^{2B4}y^{3B5} = e^{B1}e^{B2T}y^{B3+ 2B4+ 3B5}.

    So taking C1= e^{B1}, C2= B3+ 2B4+ 3B5, [tex], and C3= -B3 you will have t= C1(y^{C2})e^{-C3T}. I do not believe you can choose the constants so that it is e^{-C/T} rather than e^{-CT}.
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  3. #3
    jas
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    by
    ln t = B1 + B2.T + B3.X + B4.X^2 + B5.X^3
    i mean
    ln t = B1 + B2.T + B3.(ln y) + B4.(ln y)(ln y) + B3.(ln y)(ln y)(ln y)
    so (ln y)(ln y) is not the same as (ln y^2) = 2 ln y ?

    so i don't know if the simplification you have is true?
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