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Math Help - Can you help me with this simplifying problem

  1. #1
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    Can you help me with this simplifying problem

    I just want to confirm that the answer to this problem

    54^j / 27^j3

    54 power j power 1 / 27 j power 3

    I come up with this answer 54^j-2 or should i write it 54^j-j3. Whats the proper format?
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    Re: Can you help me with this simplifying problem

    Quote Originally Posted by Capnzims View Post
    I just want to confirm that the answer to this problem

    54^j / 27^j3

    54 power j power 1 / 27 j power 3

    I come up with this answer 54^j-2 or should i write it 54^j-j3. Whats the proper format?
    \displaystyle \begin{align*} \frac{54^j}{27j^3} &= \frac{\left(2 \cdot 27\right)^j}{27j^3} \\ &= \frac{2^j \cdot 27^j}{27j^3} \\ &= \frac{2^j \cdot 27^{j-1}}{j^3} \end{align*}
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    Re: Can you help me with this simplifying problem

    The book im using says that this answer would work too please tell me if this is correct
    2/j^2 is that wrong?
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    Re: Can you help me with this simplifying problem

    Quote Originally Posted by Capnzims View Post
    The book im using says that this answer would work too please tell me if this is correct
    2/j^2 is that wrong?
    is the original expression \frac{54^j}{27^{j^3}} or \frac{54j}{27j^3} ?
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    Re: Can you help me with this simplifying problem

    Quote Originally Posted by skeeter View Post
    is the original expression \frac{54^j}{27^{j^3}} or \frac{54j}{27j^3} ?
    Or even \displaystyle \begin{align*} \frac{54^j}{\left(27^j\right)^3} \end{align*}?
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    Re: Can you help me with this simplifying problem

    the j's are exponents raised to the power of the number beside them.
    so j to the first power and j cubed
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    Re: Can you help me with this simplifying problem

    Quote Originally Posted by Capnzims View Post
    the j's are exponents raised to the power of the number beside them.
    so j to the first power and j cubed
    then \frac{2}{j^2} would not be correct.

    however, since the "book" says it's a valid simplification, leads me to believe the original expression was meant to be \frac{54j}{27j^3}
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