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Y=KX^n Graphing and Coefficient

  1. Nov 5, 2011 #1
    So there's this graph in chemistry with a curved negative slope shape.

    (x, y)
    (0.16, 104.61)
    (0.26, 39.62)
    (0.35, 21.86)
    (0.56, 8.54)
    (0.70, 5.47)
    (0.98, 2.79)


    The equation is y=KX^N

    1) What is the value of N for which the graph would be a straight line?
    2) What is the value of K for which the graph would be a straight line?


    No idea how to do this.
     
  2. jcsd
  3. Nov 5, 2011 #2

    symbolipoint

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    Using purely qualitative understanding of functions, you would expect any equation of the form, y=kx where y is the dependant varialbe, x is the independant variable, and k is any real nonzero number constant, the be an equation for a line. Knowing that, if you wished y=kx^n to be a line, then n=1, and again, k is any real constant not equal to zero.
     
  4. Nov 6, 2011 #3

    symbolipoint

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    Your (x, y) data show x going through one cycle of 10 and y going through about two cycles of 10. An exponential fit might be reasonable to try. You get a better idea if you plot the points on cartesian paper. Maybe semilog paper, too.
     
  5. Nov 6, 2011 #4
    i tried exponentially, and if i put in logy and input y values, then its almost a straight line...
    how do i find n?
     
  6. Nov 6, 2011 #5

    I like Serena

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    Welcome to PF, lunapt!:

    Consider that:
    [itex]y=K X^N[/itex]
    [itex]\log y=\log(K X^N)[/itex]
    [itex]\log y=\log K + N \log X[/itex]

    The method would be to set up the regular equation for a line.
    And then match the coefficients with log(K) and N.

    However, it seems to me that your equation wouldn't fit (but I did not work it out).
    It seems more likely that your relation is [itex]y=K N^X[/itex].
    So:
    [itex]\log y = \log(K N^X)[/itex]
    [itex]\log y = \log K + X \log N[/itex]
     
  7. Nov 8, 2011 #6

    symbolipoint

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    lunapt,

    Your points plotted on a cartesian system directly appear like exponential decay. Your points when treated as Log(y) versus Log(x) and then plotted seem to give a very clean, accurate, line.
     
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