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Math Help - exponents

  1. #1
    Newbie lmschneider's Avatar
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    exponents

    Problem:
    (2^9/2^8 * sixth root of 3)^4

    We have to solve it labeling each step.

    I have it down to

    (2^1 * sixth root of 3)^4 by the quotient of powers

    I am stuck from there, though.
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  2. #2
    Junior Member universalsandbox's Avatar
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    (2^9/2^8 * sixth root of 3)^4

    You are getting closer.

    2^9/2^8 = 2^1 or more commonly known as 2.

    because 2^(9-8) = 2^1

    sixth root of three can be written as 3^(1/6)

    so.

    [ 2*3^(1/6) ]^4
    Apply the 4 to inside the bracket

    2^4 * 3^[ (1/6)*4) ]

    16 * 3^(4/6)
    16 * 3^(2/3)
    =33.281341
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  3. #3
    Newbie lmschneider's Avatar
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    Thank you, Darlin!
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  4. #4
    Newbie
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    Hey there lmschneider,

    It's a bit hard to establish what your question looked like from the notation, but taking it literally...

    (2^9/2^8 * sixth root of 3)^4 is (\frac {2^9}{2^8} * \sqrt[6]{3}) ^4

    If this is the case...

    (\frac {2^9}{2^8} * \sqrt[6]{3}) ^4 = (2 \sqrt[6]{3})^4

    = (2 * 3^{\frac{1}{6}})^4

    = (2^4 * 3^{\frac{4}{6}})

    = 2^4 * 3^{\frac{2}{3}}

    = 16 * \sqrt[3]{3^2}

    = 16 \sqrt[3]{9}

    Trust this helps...
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  5. #5
    Newbie lmschneider's Avatar
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    Thank you!
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