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Math Help - Arithmetic : Powers and Roots Question

  1. #1
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    Arithmetic : Powers and Roots Question

    Hi All,
    I'm new to the forum so firstly hello. The following question may be rudimentary given the level of ability of some of you but I would very much appreciate a clear understanding/answer.

    1. Can someone help me understand how 2√27 simplifies to (2 X 3 X √3).
    2. Can someone help me understand how 3^√27 X 3^√3 simplifies to 3^3√3

    I've read quite a few of the web manuals on Powers and Roots but these specific examples are a block for me. If anyone can point me towards a comprensive resource I'd very much appreciate it.

    Thanks in Advance.
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  2. #2
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    Do you understand that \sqrt{27}=\sqrt{3^2\cdot 3}=\sqrt{3^2}\cdot\sqrt{3}~?
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  3. #3
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    \displaystyle 3^{\sqrt{27}} \times 3^{\sqrt{3}} = 3^{\sqrt{27} + \sqrt{3}}.

    Can you simplify \displaystyle \sqrt{27}?
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    Quote Originally Posted by Plato View Post
    Do you understand that \sqrt{27}=\sqrt{3^2\cdot 3}=\sqrt{3^2}\cdot\sqrt{3}~?
    Hi Plato:
    Thanks for your reponse. I get that \sqrt{3^2\cdot 3}=\sqrt{3^2}\cdot\sqrt{3}~?, I don't know how one derives the first bit: \sqrt{27}=\sqrt{3^2\cdot 3}
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  5. #5
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    Because \displaystyle 9\cdot 3 = 27... You need to look for square factors. In this case, you can see that \displaystyle 3^2 = 9 is a factor.
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    Quote Originally Posted by Prove It View Post
    Because \displaystyle 9\cdot 3 = 27... You need to look for square factors. In this case, you can see that \displaystyle 3^2 = 9 is a factor.
    Removed. figured it out. So just to post the solution...

    1. Can someone help me understand how 2√27 simplifies to (2 X 3 X √3).
    1a. Factor of 27 is 3 * 9. 9 is 3^2
    1b. Simplifies to 2√3^2 X 2 = 2√3^2 * √3
    1c. √3^2 = 3
    1d. Leaving 2 * 3 * √3


    2. Can someone help me understand how 3^√27 X 3^√3 simplifies to 3^3√3
    2a. Factor of 27 is 3 * 9. 9 is 3^2
    2b. Simplifies to 3 X √3 * 9
    2c. Simplifies to 3 X √3 * √3^2
    2d. Leaving 3 X 3 * √3

    Thanks to all.
    Last edited by dumluck; January 6th 2011 at 05:11 AM.
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  7. #7
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    I would review how to get the prime factorization of a number using factor trees. For example, do you see why

    3000 = 2^3\cdot 3 \cdot 5^3?

    Now, you can take the square root of 3000 as follows:

    \sqrt{3000} =\sqrt{2^3\cdot 3 \cdot 5^3}=\sqrt{2^2\cdot 5^2}\cdot \sqrt{2\cdot 3 \cdot 5}=2\cdot 5\sqrt{30}=10\sqrt{30}

    This is a systematic procedure that will always get you the answer. After you're comfortable with this you can begin to use shortcuts such as

    \sqrt{3000} =\sqrt{100\cdot 30}=10\sqrt{30}

    Note that I know that \sqrt{30} can't be reduced further because the prime factorization of 30 has no exponents greater than 1.


    Perhaps someone who is better at latex than me can draw a factor tree for 3000?
    Last edited by DrSteve; January 6th 2011 at 05:24 AM.
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