# relationships of the form y=kx^n

• Nov 17th 2012, 06:21 AM
LoneWolf
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.
• Nov 17th 2012, 06:45 AM
skeeter
Re: relationships of the form y=kx^n
$log{y} = \log(kx^n)$

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

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

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

$y = b + mx$

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

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

$y = -1.7 + 4x$

which translates to

$y = 0.02 \cdot x^4$
• Nov 19th 2012, 09:43 AM
LoneWolf
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)

$y = -1.7 + 4x$

which translates to

$y = 0.02 \cdot x^4$

That is the part where I don't understand, what did you mean by y = –1.7 + 4x translates to y = 0.02x^4 ?
• Nov 19th 2012, 10:50 AM
skeeter
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 ?

$y = kx^n$

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

$n = 4$