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Power and Exponential Equations

  1. Jun 15, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm writing a lab report and I'm having trouble understanding how to set up these experimental equations.

    I understand Linear (y=mx+b) but my professor wants us to use separate equations to illustrate power and exponential forms.
    Power : y=bx^m
    Exponential: y=be^(mx)

    On the first lab report for the power relationship I had:
    Slope of linearization equation (trendline) = 4
    Y intercept of linearization equation = .9031
    I set up my equation as y=.9031x^4

    For the exponential relationship I had:
    Slope of linearization equation (trendling) = 1.3029
    Y intercept of linearization equation (trendline) = .699
    I set up my equation as y=.699e^(1.3029x)

    Does anyone know how to use these equations?

    2. Relevant equations

    Power : y=bx^m
    Exponential: y=be^(mx)

    3. The attempt at a solution

    y=.9031x^4
    y=.699e^(1.3029x)


    Thanks!
     
  2. jcsd
  3. Jun 15, 2011 #2

    ojs

    User Avatar

    Hmm, I am sorry, I don't understand your work here.

    Are you using data sets that show linear relationship, making a linear fit to them and using the parameters from that fit to put into a power and exponential equation? That won't work at all.

    If you have data sets that show either power or exponential relations then you either use a program to fit that data to the given equation or take the n-th root or logarithm of the y-data and then use linear fitting.

    And what do you mean by "how to use these equations"?
     
  4. Jun 15, 2011 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    The claim in your first lab report looks wrong. If you say "slope of trendline = 4" (after your word "linearization") you are claiming an equation of the form y = a + 4*x---that is what we mean when we say a linear function with trendline having slope 4. However, if you were looking at *log(y)* vs. *log(x)* and getting a linear form with trendline = 4 (that is, log(y) = c + 4*log(x)), _then_ you would, indeed, have y = a*x^4.

    RGV
     
  5. Jun 15, 2011 #4
    Thanks I think I got it
     
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