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Math Help - relationships of the form y=kx^n

  1. #1
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    Question 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 Thanks.
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  2. #2
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    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
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    Re: relationships of the form y=kx^n

    Quote Originally Posted by skeeter View Post

    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 ?
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  4. #4
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    Re: relationships of the form y=kx^n

    Quote Originally Posted by LoneWolf View Post
    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
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