Y=KX^n Graphing and Coefficient

  • Thread starter lunapt
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  • #1
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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.
 

Answers and Replies

  • #2
symbolipoint
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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.
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.
 
  • #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.
 
  • #4
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i tried exponentially, and if i put in logy and input y values, then its almost a straight line...
how do i find n?
 
  • #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]
 
  • #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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