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Math Help - kinda hard exponent problem

  1. #1
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    kinda hard exponent problem

    q23 (answer 4)

    Given 2x^2 = 3y^3

    By how much will y increase if x is multiplied by 2?

    1) 2^(1/3)
    2) 4^(1/3)
    3) 8


    kinda hard exponent problem-exponentq23.jpg

    thanks in advance!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: kinda hard exponent problem

    Reform the given equation to y=...
    Afterwards replace x by 2x.
    Then you'll see y is increased by a factor ... ?
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  3. #3
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    Re: kinda hard exponent problem

    y= (2/3x^2)^(1/3)
    2^2=4
    4*2/3 = 8/3
    8/3^(1/3)
    isn't an answer option... grrr
    that 3 keeps bugging me out!
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: kinda hard exponent problem

    What I mean is the following:
    2x^2=3y^3
    \Leftrightarrow y=\sqrt[3]{\frac{2x^2}{3}} (1)
    Because the question asks how y increases if x is multiplied by two you have to replace x by 2x
    \Leftrightarrow y=\sqrt[3]{\frac{2(2x)^2}{3}}
    \Leftrightarrow y=\sqrt[3]{\frac{2\cdot 4x^2}{3}}
    \Leftrightarrow y=\sqrt[3]{4}\cdot \sqrt[3]{\frac{2x^2}{3}} (2)

    So you recognize an equal part in (1) and (2), but (2) is multiplied by a factor ... ?
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  5. #5
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    Re: kinda hard exponent problem

    I thought I read The problem wrong. That's why I included the image. Again I don't know the answer. I got the same reasoning solution as you
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  6. #6
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    e^(i*pi)'s Avatar
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    Re: kinda hard exponent problem

    Quote Originally Posted by jsvlad View Post
    I thought I read The problem wrong. That's why I included the image. Again I don't know the answer. I got the same reasoning solution as you
    If you understand how Siron got that answer then use your laws of exponents to say that \sqrt[3]{4} = 4^{1/3}


    Edit: The question is asking to find the difference in "scale" if you will. In other words it wants you to find \dfrac{f(2x)}{f(x)} where f(x) is the equation given.

    From Siron's post the answer is eq2/eq1
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