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  1. #1
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    Unhappy Please Help

    I have to learn how do roots without a calculator and one of the practice problems i get was 12 root of 64 equal square root of two.How could i figure out problems like these without a calculator?Also what is (n^3m^9)^1/2?
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  2. #2
    Jen
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    Quote Originally Posted by anime_maniac20 View Post
    I have to learn how do roots without a calculator and one of the practice problems i get was 12 root of 64 equal square root of two.
    Start by factoring 64 down to it's prime factorization,
    Please Help-factorization-64.bmp

    So 64 can be written as $\displaystyle 2^6$

    Keep this in mind.

    So by the definition of roots, $\displaystyle \sqrt[12]{64}=64^{\frac{1}{12}}$

    Now substituting the prime factorization of 64 in place of 64, and writing it as an exponent as previously shown we get,

    $\displaystyle \left(2^6\right)^{\frac{1}{12}}$

    Using the law for the power of a power, we multiply the exponents together which gives us,

    $\displaystyle 2^{\frac{6}{12}}=2^{\frac{1}{2}}=\sqrt{2}$


    Also what is (n^3m^9)^1/2?
    $\displaystyle \left(n^3m^9\right)^{\frac{1}{2}}$

    Using the same law of exponents the outside exponent gets "distributed" to each inside term.

    $\displaystyle n^{3*\frac{1}{2}}m^{9*\frac{1}{2}}=n^{\frac{3}{2}} m^{\frac{9}{2}}=\sqrt{n^3}\sqrt{m^9}$

    Or we can also do it this way,

    $\displaystyle \sqrt{n^3m^9}=\sqrt{n^3}\sqrt{m^9}$

    These are all equivalent statements.
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  3. #3
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    Using the same law of exponents the outside exponent gets "distributed" to each inside term.



    Or we can also do it this way,

    I had the exact answers but the answer that was correct was like $\displaystyle nm^{4}\sqrt{nm}$
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  4. #4
    Jen
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    Quote Originally Posted by anime_maniac20 View Post
    I had the exact answers but the answer that was correct was like $\displaystyle nm^{4}\sqrt{nm}$
    Ah...

    That is correct, there was one step of simplification that I missed, thank you.

    We can further simplify this by,



    $\displaystyle
    \sqrt{n^3m^9}=\sqrt{n^3}\sqrt{m^9}=\sqrt{n^2*n}\sq rt{m^8*m}=nm^4\sqrt{nm}
    $

    I skipped a few steps there where I used the rules I previously discussed. Let me know if you follow this o.k.
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  5. #5
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    Thank you so much,I did understand it. You mad it look really simple.
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