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Math Help - [SOLVED] Simplifying Rational Exponents

  1. #1
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    [SOLVED] Simplifying Rational Exponents

    I need to simplify:
    (49c^6)^(-3/2)
    I know that the answer is 1/343c^9. But I have noo idea how they got it?
    Thank you!
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  2. #2
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    (49c^{6})^{\frac{-3}{2}}

    \frac{1}{(49c^{6})^{\frac{3}{2}}}

    \frac{1}{(49^{3}c^{18})^{\frac{1}{2}}}

    \frac{1}{343c^{9}}
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  3. #3
    Senior Member topher0805's Avatar
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    (49c^6)^{-\frac {3}{2}}

    Recall the rule of negative exponents:

    x^{-a} = \frac {1}{x^a}

    So we have:

    <br />
(49c^6)^{-\frac {3}{2}} = \frac {1}{(49c^6)^{\frac {3}{2}}}

    Also recall that:

    x^{\frac {a}{b}} = (x^{\frac {1}{b}})^a

    So we have that:

    \frac {1}{(49c^6)^{\frac {3}{2}}} = \frac {1}{\left((49c^6)^{\frac {1}{2}}\right)^3}

    Recall that:

    x^{\frac {1}{2}} = \sqrt {x}

    This gives us:

    \frac {1}{\left((49c^6)^{\frac {1}{2}}\right)^3} = \frac {1}{(7c^3)^3} = \frac {1}{343c^9}
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  4. #4
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    thank you soo much topher0805!! and galactus!
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  5. #5
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    Ok, I understand what happened with my 1st question... now i'm stuck on another
    How does (4^-2)+(3^-1)/(3^-2)+(2^-3) become 1(23/34)?
    This is waht I'm doing:

    = (1/4^2+3^-1)/(1/3^2+2^3)
    = (1/16+3)/(1/9+8)
    = (1/19)/(1/17)
    = (1/19)*(17/1)
    = 323.
    I have no idea where I am going wrong. thank you for any tips you might have!
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ria3 View Post
    Ok, I understand what happened with my 1st question... now i'm stuck on another
    How does (4^-2)+(3^-1)/(3^-2)+(2^-3) become 1(23/34)?
    This is waht I'm doing:

    = (1/4^2+3^-1)/(1/3^2+2^3)
    = (1/16+3)/(1/9+8)
    = (1/19)/(1/17)
    = (1/19)*(17/1)
    = 323.
    I have no idea where I am going wrong. thank you for any tips you might have!
    Two things.
    1) New questions should go in new threads.

    2) Use parenthesis! This is a mess. Is your problem
    \frac{4^{-2} + 3^{-1}}{3^{-2} + 2^{-3}}

    If so then this is
    = \frac{\frac{1}{16} + \frac{1}{3}}{\frac{1}{9} + \frac{1}{8}}

    You appear to be losing sight of which fractions are which because you aren't using the parenthesis properly.

    -Dan
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  7. #7
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    k, sorry, I'll make an new thread next time.
    thank you for the help topsquark!
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