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Math Help - Exponential modelling

  1. #1
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    Exponential modelling

    Ok, I know how to do this but I don't have a graphing calculator now so. Question is:
    (2, 13.2) (27, 40.5) (58, 82.7) (82, 172), find exponential function f(x) = ba^x
    Last edited by mr fantastic; September 6th 2009 at 07:09 PM. Reason: Re-titled the post
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  2. #2
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    Quote Originally Posted by SkyHigh View Post
    Ok, I know how to do this but I don't have a graphing calculator now so. Question is:
    (2, 13.2) (27, 40.5) (58, 82.7) (82, 172), find exponential function f(x) = ba^x
    Choose two points to substitute into the equation.

    Then you'll have two equations in two unknowns to solve simultaneously.


    In your case...

    13.2 = ba^2

    40.5 = ba^{27}


    Notice that ba^{27} = ba^2a^{25} = 13.2a^{25}


    So 40.5 = 13.2a^{25}

    \frac{40.5}{13.2} = a^{25}

    \left(\frac{40.5}{13.2}\right)^{\frac{1}{25}} = a.


    I'm sure you can solve for b.
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    Quote Originally Posted by SkyHigh View Post
    Ok, I know how to do this but I don't have a graphing calculator now so. Question is:
    (2, 13.2) (27, 40.5) (58, 82.7) (82, 172), find exponential function f(x) = ba^x
    Note that \log y = \log (b a^x) = \log b + x \log a.

    So my advice is to take plot the points ( \log y, ~ x) and draw the line of best fit through them. The gradient of this line can be used to get your best estimate of a and the \log y-intercept can be used to get your best estimate of b.
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    Quote Originally Posted by SkyHigh View Post
    Ok, I know how to do this but I don't have a graphing calculator now so. Question is:
    (2, 13.2) (27, 40.5) (58, 82.7) (82, 172), find exponential function f(x) = ba^x
    4 points, 2 parameters. Are you doing a curve fitting?
    Last edited by mr fantastic; September 18th 2009 at 08:17 AM. Reason: Restored original reply
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    Prove it, your method doesn't account for the last two points so it's not right I tried it.
    This should be a very easy problem if I plot the points in the graphing calculator and find Exp Reg. Please someone with a graphing calculator kindly do it for me I don't have one.
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    Quote Originally Posted by SkyHigh View Post
    Prove it, your method doesn't account for the last two points so it's not right I tried it.
    This should be a very easy problem if I plot the points in the graphing calculator and find Exp Reg. Please someone with a graphing calculator kindly do it for me I don't have one.
    I outlined how to do it by hand in my earlier post. If you've been given a question like this it's expected you have access to the necessary technology. If you do a Google search I'm sure you'll find an on-line regression applet suitable for your purposes.
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    Thanks I don't know there's a thing a called Regression Applet before. I got my answer using it (Exponential Regression Applet).
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    Quote Originally Posted by SkyHigh View Post
    Thanks I don't know there's a thing a called Regression Applet before. I got my answer using it (Exponential Regression Applet).
    I assumed the points lay on the curve, I didn't realise you were using a curve of best fit...
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