relationships of the form y=kx^n

I have a set of data:

x 1 2 3 5 10

y 0.02 0.32 1.62 12.53 199.80

And I am told that the suspected relationship between x and y is of the form y = kx^n

I need to explain how a graph of log(x) against log(y) tells me whether this is a good model for relationship? I know that taking logs of x and y and plotting them on a graph will give me a straight line, but how does this tell me that the model (y=kx^n) is appropriate for the relationship?

P.S. What does the straight line of log x and log y shows? Is it like a line of best fit?

Would be grateful for any relevant explaination or a hint, I am trying to figure this out for the past 2 hours (Worried) Thanks.

Re: relationships of the form y=kx^n

$\displaystyle log{y} = \log(kx^n)$

$\displaystyle \log{y} = \log{k} + \log{x^n}$

$\displaystyle \log{y} = \log{k} + n\log{x}$

now compare each term above to the slope-intercept form of a linear equation ...

$\displaystyle y = b + mx$

the y-intercept is $\displaystyle \log{k}$ and the slope is $\displaystyle n$, the exponent of your power function.

using your data, I get an logarithmic linear regression of about (coefficients rounded)

$\displaystyle y = -1.7 + 4x$

which translates to

$\displaystyle y = 0.02 \cdot x^4$

Re: relationships of the form y=kx^n

Quote:

Originally Posted by

**skeeter**

using your data, I get an logarithmic linear regression of about (coefficients rounded)

$\displaystyle y = -1.7 + 4x$

which translates to

$\displaystyle y = 0.02 \cdot x^4$

That is the part where I don't understand, what did you mean by y = –1.7 + 4x translates to y = 0.02x^4 ?

Re: relationships of the form y=kx^n

Quote:

Originally Posted by

**LoneWolf** That is the part where I don't understand, what did you mean by y = –1.7 + 4x translates to y = 0.02x^4 ?

$\displaystyle y = kx^n$

$\displaystyle k = 10^{-1.7} = .02$

$\displaystyle n = 4$