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Math Help - Rational Exponents

  1. #1
    Ash
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    Rational Exponents

    I need help in learning how to do these two problems.

    #1 (3^1/2 2^3/2)^2

    #2 b^2/5 (b^8/5 + b 3/5 ) *Assuming that b is positive

    I know the answer to #1 is 24 but, the only way I obtained a 24 was by:

    (3^1/2 2^3/2)^2
    3^1 2^3
    (3) 8 = 24, but, I thought, you're suppose to square the 3 & 2 and change it into (9^1 & 4^3) I'm confused by both problems.

    Can you show me both steps so I can see where I am going wrong and how to obtain the answers.
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  2. #2
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    Quote Originally Posted by Ash View Post
    I thought, you're suppose to square the 3 & 2 and change it into (9^1 & 4^3)
    Why would you use the exponent twice? You DID use the outside '2' when you modified the internal exponents.

    [3^(1/2)]^2 = 3^1

    [2^(3/2)]^2 = 2^3

    There is no more 2 in the exponent.
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  3. #3
    Ash
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    Quote Originally Posted by TKHunny View Post
    Why would you use the exponent twice? You DID use the outside '2' when you modified the internal exponents.

    [3^(1/2)]^2 = 3^1

    [2^(3/2)]^2 = 2^3

    There is no more 2 in the exponent.
    Thanks, that makes sense, Do you know the steps to solving the second one?
    Last edited by Ash; September 16th 2007 at 04:06 PM. Reason: spelling
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  4. #4
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    You have the rules, don't you?

    a^{b}*a^{c} = a^{b+c}

    That's all it takes.

    2/5 + 8/5 = 10/5 = 2
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  5. #5
    Ash
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    Quote Originally Posted by TKHunny View Post
    You have the rules, don't you?

    a^{b}*a^{c} = a^{b+c}

    That's all it takes.

    2/5 + 8/5 = 10/5 = 2
    So you first add the 2/5 + 8/5 = 2, then add the 2/5 + 3/5 =1

    So, the answer is, b^2 + b



    Thanks again for your help.
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  6. #6
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    yeah you just need to study it , its really easy
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