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Math Help - Raised by an unknown quantity

  1. #1
    Newbie PxEckx's Avatar
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    Question Raised by an unknown quantity

    Hi! Here I'm again!

    If 5^{3a} = 64 , so 5^{-a} is?

    I've tried change the base, (  64 = 2^6 ), for example... But I coudn't get a result, because I don't know what to do with the 5^3 = 125 and the 5^a , or how to transform they on 2

    >>>

     <br />
5^{3a} = 64

    5^3\times5^a = 2^6<br />

    This is the way?
    How can I do this?!
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by PxEckx View Post
    Hi! Here I'm again!

    If 5^{3a} = 64 , so 5^{-a} is?

    I've tried change the base, (  64 = 2^6 ), for example... But I coudn't get a result, because I don't know what to do with the 5^3 = 125 and the 5^a , or how to transform they on 2

    >>>

     <br />
5^{3a} = 64

    5^3\times5^a = 2^6<br />

    This is the way?
    How can I do this?!
    Hi PxEckx,

    \displaystyle \left(5^{-a}\right)^{-3}=5^{3a}=64

    \displaystyle \left(\left(5^{-a}\right)^{-3}\right)^{-\frac{1}{3}}=\left(64\right)^{-\frac{1}{3}}

    \displaystyle 5^{-a}=\left(64\right)^{-\frac{1}{3}}=\text{?}
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  3. #3
    Newbie PxEckx's Avatar
    Joined
    Jun 2010
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    SP, Brasil
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    I forgot this 'peculiarity'! Perfect! Rise both sides by the same number!!!

    So:::

    (5^{-a})^{-3} = 5^{3a} = 64

    (5^{3a})^-\displaystyle^\frac{1}{3} = (64)^-^\frac{1}{3}

    5^{-a} = \displaystyle\frac{1}{\sqrt[3]{64} }

    5^{-a} = \displaystyle\frac{1}{4}

    Thanx!!!
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