# Y=KX^n Graphing and Coefficient

1. Nov 5, 2011

### lunapt

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. Nov 5, 2011

### symbolipoint

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. Nov 6, 2011

### symbolipoint

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. Nov 6, 2011

### lunapt

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. Nov 6, 2011

### I like Serena

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

6. Nov 8, 2011

### symbolipoint

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.