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Using logs to obtain the slope of a function

  1. Jan 26, 2007 #1
    1. The problem statement, all variables and given/known data
    I have to find the slope of an exponential curve by using log identities.

    Using basic values such as:
    (10, 0.692)
    (30, 1.082)
    (70, 1.579)

    2. Relevant equations
    The equation is in the form T=Cm^p
    So the log function is logT=logC + Plog(m) right?

    3. The attempt at a solution

    When I take the log of (10, 0.692) I end up with negative values, which is completely non-sensical when compared to my experimental values!
    I end up with function -.6x^.4
    when I should get .3x^.4
  2. jcsd
  3. Jan 27, 2007 #2
    To me there is still a fair bit of ambiguity in the statement of your problem, perhaps if you could elaborate a bit further, it would be easier to assist you. For example, the ordered pair (10, 0.692), what values does this correspond to in the equation? Your formula is certainly right, but I'm not quite following. I assume the abscissa is Temperature, but what's the other?
  4. Jan 27, 2007 #3


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    Science Advisor

    Why is it non-sensical? you are talking about the logarithm being negative, not the values themselves.

    I'm going to assume that the first number in each pair is m, the second number T. As Kreizhn said, it would be good idea to actually put that in your question.

    (log 10, log 0.692)= (1, -0.160)
    (log 30, log 1.082)= (1.477, 0.0342)
    (log 70, log 1.579)= (1.845, 0.198)

    slope, P:
    (0.198-(-1.60))/(1.845- 1)= 0.358/0.845= 0.424
    When m= 10, log 0.692= -.160= log C+ (0.424)log 10= log C+ 0.424 so
    log C= -.160- 0.424= -.584 which means C itslelf is less than 1 but positive:
    log T= -.584+ 0.424log m so T= 10-.584 m0.424= 0.2550m0.424. Rounded to one decimal place, that is T= .3 m.4.

    log C is -.6, C= .3 (again, to one decimal place)
    Last edited by a moderator: Jan 27, 2007
  5. Jan 27, 2007 #4
    That was a flawless answer. Thank you very much.
    I'm sorry for wording my question so poorly.
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