1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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.

  2. jcsd
  3. Apr 11, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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


    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):


    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:

  6. Apr 11, 2012 #5


    User Avatar
    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:




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


    User Avatar

    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.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook