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Thread: Find an approimate equation...

  1. #1
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    Find an approimate equation...

    Find an approximate equation $\displaystyle y=ab^{x}$ of the exponential curve that contains the given pair of points. Round the value of b to two decimal places.

    (0, 256) and (7, 23)

    my work:

    $\displaystyle 256b^7=23$


    $\displaystyle 256^7=\frac{23}{256}$


    $\displaystyle b=\sqrt[7]{\frac{23}{256}}$ which is $\displaystyle b=4.7$


    Thus, $\displaystyle y=256(4.7)^{x}$


    Is this the right answer?
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  2. #2
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    Quote Originally Posted by mt_lapin View Post
    Find an approximate equation $\displaystyle y=ab^{x}$ of the exponential curve that contains the given pair of points. Round the value of b to two decimal places.

    (0, 256) and (7, 23)

    my work:

    $\displaystyle 256b^7=23$


    $\displaystyle 256^7=\frac{23}{256}$


    $\displaystyle b=\sqrt[7]{\frac{23}{256}}$ Mr F says: Correct.

    which is $\displaystyle b=4.7$ Mr F says: Time to get a new calculator. The seventh root of 23/256 cannot possibly be greater than 1! I get 0.71, correct to two decimal places.


    Thus, $\displaystyle y=256(4.7)^{x}$


    Is this the right answer?
    ..
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  3. #3
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    My calucator isn't busted, I am.

    I did $\displaystyle \left(\frac{23}{256}\right)^7$ when I should have done $\displaystyle \left(\frac{23}{256}\right)^{1/7}$ ... I made a mistake.

    Thank you for checking my work!
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