Y=KX^n Graphing and Coefficient

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    Coefficient Graphing
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Discussion Overview

The discussion revolves around the equation y=KX^N and its graphical representation based on provided data points. Participants explore the conditions under which this equation would yield a straight line, considering both theoretical and practical approaches to graphing and fitting the data.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant questions the values of N and K that would make the graph a straight line, expressing uncertainty about how to approach the problem.
  • Another participant suggests that for y=KX^N to represent a straight line, N must equal 1, while K can be any nonzero constant.
  • A different perspective proposes that the data might fit an exponential model better, recommending the use of Cartesian or semilog paper for plotting the points.
  • One participant mentions that plotting log(y) against log(x) results in a nearly straight line, indicating a potential logarithmic relationship.
  • Another participant provides a transformation of the original equation into logarithmic form, suggesting that the relationship might be y=K N^X instead, although they express doubt about the fit of the original equation.
  • It is noted that the plotted points appear to exhibit exponential decay when viewed in Cartesian coordinates.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate model for the data, with some advocating for a linear interpretation under specific conditions, while others suggest an exponential or logarithmic fit may be more suitable. The discussion remains unresolved regarding the best approach to model the data accurately.

Contextual Notes

There are limitations in the assumptions made about the relationships between the variables, and the discussion does not resolve the mathematical steps necessary to determine the values of N and K definitively.

lunapt
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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.
 
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lunapt said:
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 dependent varialbe, x is the dependent 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.
 
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.
 
i tried exponentially, and if i put in logy and input y values, then its almost a straight line...
how do i find n?
 
Welcome to PF, lunapt!:

Consider that:
y=K X^N
\log y=\log(K X^N)
\log y=\log K + N \log X

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 y=K N^X.
So:
\log y = \log(K N^X)
\log y = \log K + X \log N
 
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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