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Math Help - 8.24 is not like 1 + 3x

  1. #1
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    Wink 8.24 is not like 1 + 3x

    Hi guys,

    now this is an easy one.

    Given the expansion of (1 + 3x)^{1/3} = 1 + x - x^2 + \frac{5}{3} x^3

    find (8.24)^{1/3}

    so the problem is to get 8.24 into the form 1 + 3x i.e. x = 2.413 but it can't be that! Since the index of the expansion is 1/3 the binomial is only valid if mod x < 1/3.

    Clearly I am missing something obvious!
    Last edited by mr fantastic; June 20th 2009 at 12:48 AM. Reason: Fixed superscript in latex
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  2. #2
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    Hi

    (8.24)^{\frac{1}{3}} = (8\times 1.03)^{\frac{1}{3}} = 2\cdot (1+3\cdot 0.01)^{\frac{1}{3}}
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  3. #3
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    Ah! That's it! Thankyou my friend - a very quick reply!
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  4. #4
    Super Member malaygoel's Avatar
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    8.24^{\frac{1}{3}}
    =(8+.24)^{\frac{1}{3}}
    =2(1+\frac{.24}{8})^{\frac{1}{3}}
    =2(1+3\frac{.08}{8})^{\frac{1}{3}}
    =2[1+3(.01)]^{\frac{1}{3}}
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  5. #5
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    mr fantastic's Avatar
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    Quote Originally Posted by s_ingram View Post
    Hi guys,

    now this is an easy one.

    Given the expansion of (1 + 3x)^{1/3} = 1 + x - x^2 + \frac{5}{3} x^3

    find (8.24)^{1/3}

    so the problem is to get 8.24 into the form 1 + 3x i.e. x = 2.413 but it can't be that! Since the index of the expansion is 1/3 the binomial is only valid if mod x < 1/3.

    Clearly I am missing something obvious!
    From the given expression it follows that (8 + 24x)^{1/3} = 2 (1 + 3x)^{1/3} = 2\left(1 + x - x^2 + \frac{5}{3} x^3\right) .

    Substitute x = \frac{1}{100}.
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