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Math Help - Interesting exponential/logarithm example

  1. #1
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    Interesting exponential/logarithm example

    One of the more budding students in my class derived this example question, I think his dad works in a hostpital, and its based on actual paperwork on MRSA in hospitals...anyway, If anyone can give me a hand as Im useless with exponentials..

    - Hostpital rooms can be decontaminated of MRSA if sprayed with hydrogen peroxide, however this is harmful to people if there's prolonged exposure, so we must know its decay rate.
    - A room was sprayed with vapour containing hydrogen peroxide, with a concentration of 22.80 mmoles/litre, and the level of this peroxide, ct, was measured over intervals:

    t (min)----------0--------5-------10-------20------
    Ct (mmoles/ltr) 22.80 -- 17.60 -- 13.80 -- 8.25

    As far as I can conclude, the decay equation is ct = coe^-λt

    a) how do you tranform the equation so the straight line of y=mx+c can be plotted and how are y, m and c related to ct, co, and λ
    Last edited by bobchiba; March 13th 2008 at 08:18 PM.
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  2. #2
    Moo
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    Hello,

    C_t=C_0 e^{- \lambda t}

    Generally, for getting the linear form, you transform all this stuff in logarithms.
    The graphic will be ln(Ct)=f(t), so the ordinate will be ln(Ct) and the abscissa t.

    To transform it, compose with the logarithm :

    \ln(C_t) = \ln(C_0 e^{- \lambda t})

    But we know that \ln(ab) = \ln(a) + \ln(b) and that \ln(e^x) = x

    So \ln(C_t) will be \ln(C_0) - \lambda t
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    Thanks MOO, although I admit I probably sound like I know more than I do. I grasp the concept of what you've layed down, but I don't really have much experience of going through the motions, If anyone could develop on whats there or have any other examples, would be great,
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    Moo
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    Well,

    What you can remember is that the logarithm transforms a product into a sum, because of the ln(ab)=ln(a)+ln(b) propriety =)

    You want something in the y=mx+c form. Here, t is x. y will be ln(Ct), not Ct as you can think on first sight. It's common, you have to remember that usually, you will draw a line with ln(Ct) in ordinates and t in abscissa.

    \ln(C_t) = \ln(C_0) + \ln(e^{- \lambda t}) = \ln(C_0) - \lambda t
    This is a line i didn't put previously, but it explains all, am sorry, but i don't really see what i can explain more :/

    y=mx+c

    Here, m = - \lambda and c is \ln(C_0)
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    how would I find the value of the decay constant...ie, λ
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    Moo
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    With the values given in the text

    You've been given t and C_t, so you just need two equations to find C_0 and \lambda
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