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Math Help - [SOLVED] Simplifying/Solving Rational Exponents

  1. #1
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    [SOLVED] Simplifying/Solving Rational Exponents

    I'm having trouble simplifying this:
    (2x^-2)^2(3x^4)^-3
    the answer is 4/27x^16
    but I have no idea how they got this??
    Also, solving for 3a^(4/5)=48 could you please guide me through the steps?
    thank you so much!
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  2. #2
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    sorry, forgot to show my steps so far:
    for the first qu:
    I have it up to: 1/(4x^4)(-27x^12)
    I don't understand how they got a 4 over 27x^16? I would think it would be a 1 over 27x^16.

    For the second question:
    I have 5sqrt(3a^4) = 48
    and then I'm sorta stuck... could it be: 5*4=20, so (a^20)/3 = 16?
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  3. #3
    Junior Member teuthid's Avatar
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    exponents

    use the exponent laws \left(x^{a}\right)^{b}=x^{ab} and (xy)^{a}=x^{a}y^{b}:

    \left(2x^{-2}\right)^{2}\left(3x^4\right)^{-3}=2^2x^{-4}3^{-3}x^{-12}

    then remember that x^{a}x^{b}=x^{a+b}:

    =2^2\cdot 3^{-3}x^{-16}

    now recall that x^{-a}=\left(\frac{1}{x^{a}}\right). Also note that 3^{-3} is a positive number:

    =2^2\left(\frac{1}{3^3}\right)\left(\frac{1}{x^{16  }}\right)=\frac{4}{27x^{16}}
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  4. #4
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    thank you tuephid!
    and nvm 'bout the 2nd one... i got it
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