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I have a exponetial graph but how do I calculate results?

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    I have a set of data by HPLC that will only fit in an exponential plot in excel.

    If I was to use a linear graph I would use y=mx+c but the data will NOT fit into a linear plot.

    2. Relevant equations

    The R2 for the exponential graph = 0.9973

    y=0.0044e(0.0017x)

    where (0.0017x) is superscript.

    3. The attempt at a solution

    I have a value for my recovery that I can read off the graph but don't know how to convert this to ug (I can do this if the graph is linear).

    I have plotted µg vs peak response.

    Thanks!
     
  2. jcsd
  3. Apr 11, 2012 #2

    SammyS

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    Staff Emeritus
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    Hello Newbie_ . Welcome to PF !

    Take the natural logarithm, Ln, of both sides of y=0.0044e(0.0017x) .

    If you graph Ln(y) versus x, the graph should be a straight line.
     
  4. Apr 11, 2012 #3
    OK thanks for the info + welcome.

    I have plotted concentraion (µg) vs ln response and the line is linear.

    However I'm a bit confused about reading results from the graph ie

    y=0.0017x-5.4289

    My ln of the value I want to read off the graph = -2.57286965




    If I did not use ln I would (although the values would not be -ve):

    (-2.57286965-(-5.4289))/.0017




    How do I calculate given that I have used ln???

    ie when do I do the anti-ln? Or is the above calculation correct as I have taken the ln of all standard values + am using the ln of the recovery value?

    I hope this makes sense!

    Thanks again.
     
    Last edited: Apr 11, 2012
  5. Apr 11, 2012 #4
    I think this is correct as I have just put in a theoretical example and the "right" result came out. I could be wrong though :eek:

    Thanks
     
  6. Apr 11, 2012 #5

    Mentallic

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    Homework Helper

    If you have an exponential function of the form [tex]y=Ae^{kx}[/tex] for some constants A and k, then taking the log of both sides gives [tex]\ln(y)=\ln(Ae^{kx})[/tex][tex]=\ln(A)+\ln(e^{kx})[/tex][tex]=kx+\ln(A)[/tex]

    So what we have plotted is [tex]y'=kx+\ln(A)[/tex] where [itex]y'=\ln(y)[/itex]

    Now, the graph you have in your example is [itex]y'=0.0017x-5.4289[/itex] so if you want to find the y value of, say, at x=2, then plug x=2 into the equation to get [itex]y'=-5.4255[/itex] and since [itex]y'=\ln(y)[/itex] then [itex]e^{y'}=y[/itex] thus we get [itex]y=e^{-5.4255}=0.00440[/itex]

    If we want it the other way around, that is to find the x value when y=3 for example, we can plug y=3 into the equation and re-arrange to solve for x, or we can re-arrange the equation to make x the subject right off the bat:

    [tex]\ln(y)=kx+\ln(A)[/tex]

    [tex]kx=\ln(y)-\ln(A)[/tex]

    [tex]x=\frac{\ln\left(\frac{y}{A}\right)}{k}[/tex]

    So for y=3, A=0.0044, k=0.0017, we have [itex]x\approx 3838[/itex]
     
  7. Apr 11, 2012 #6

    NascentOxygen

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    Staff: Mentor

    It might be clearer if you plotted your data on log-scaled graph paper. You can design and print off a sample at this site: http://incompetech.com/graphpaper/logarithmic/

    It can produce both log-linear and log-log. (I thought you'd need log-linear, for ex to generate a straight line.)
     
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