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Math Help - Help!! Multiplication of Radicals

  1. #1
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    Help!! Multiplication of Radicals

    Need help please:

    1.) -2 4^√2 . 3^√3

    2.) √10 . 2 3^√100

    3.) 3^√7 . 4^√7

    4.) 3^√a^2 . 7^√a^6

    5.) 5 4^√a^3 . 5^√a^4

    6.) 3^√x^2y . √xy^3

    7.) 2 3^√16 . 6^√4

    8.) 6 3^√xy^2 . √xy

    9.) 5 3^√a . 4√a^3

    10.) 5 3^√a^2 . 7 5^√a^6


    Note:

    3^√ denotes cube root, 4^√ denotes fourth root and so on.........




    thank you =)
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  2. #2
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    Quote Originally Posted by rexII143 View Post
    Need help please:

    1.) -2 4^√2 . 3^√3

    2.) √10 . 2 3^√100

    3.) 3^√7 . 4^√7

    Note:

    3^√ denotes cube root, 4^√ denotes fourth root and so on.........




    thank you =)
    Hope I understand your notation correctly.

    1.) -2*4^(1/4)*3^(1/3)

    -2*sqrt(2)*3^(1/3)

    2.) sqrt(10)*2*100^(1/3)

    sqrt(10)*2*10^(2/3)

    10^(1/2)*2*10^(2/3)

    20*10^(1/6)

    3.) 7^(1/3)*7^(1/4)

    7^(7/12)

    And I'll complete the others later during my lunch break- let me know if that's the notation you intended.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rexII143 View Post
    Need help please:

    1.) -2 4^√2 . 3^√3
    Typically I would take this to be simplified. I'm only going to do the first one to check that I am showing you how to do what is intended. I am presuming you want this all to be put under one radical.

    There are two basic concepts here:
    1) \sqrt[n]{a} = (a)^{1/n}

    2) a^{(1/n)} b^{(1/n)} = (ab)^{(1/n)}

    (Those and getting the lcm of two numbers.)

    So let's simplify things a bit and just look at the \sqrt[4]{2} \cdot \sqrt[3]{3}.

    \sqrt[4]{2} \cdot \sqrt[3]{3} = 2^{1/4} \cdot 3^{1/3}

    Now, take a look at the denominators of the fractional exponents: 4 and 3. we want the least common multiple of 4 and 3. This is 12. So:
    \sqrt[4]{2} \cdot \sqrt[3]{3} = 2^{1/4} \cdot 3^{1/3}  = 2^{3/12} \cdot 3^{4/12} = (2^3)^{1/12} \cdot (3^4)^{1/12} = (2^3 \cdot 3^4)^{1/12}

     = (8 \cdot 81)^{1/12} = \sqrt[12]{648}

    Let me know if this is the kind of thing you were after.

    -Dan
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  4. #4
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    ^^ yes, that's it.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rexII143 View Post
    ^^ yes, that's it.
    Um, me (topsquark) or AfterShock? We both have valid and different ways of looking at it.

    -Dan
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  6. #6
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    Hello, Rex!

    I'll do a few more of them . . .


    5)\;\;5\sqrt[4]{a^3}\cdot\sqrt[5]{a^4}

    We have: . 5\cdot a^{\frac{3}{4}}\cdot a^{\frac{4}{5}} \;=\;5\cdot a^{(\frac{3}{4}+\frac{4}{5})} \;= \;\boxed{5\cdot a^{\frac{31}{20}}}



    6)\;\;\sqrt[3]{x^2y}\cdot\sqrt{xy^3}

    We have: . \left(x^2y\right)^{\frac{1}{3}}\left(xy^3\right)^{  \frac{1}{2}} \;=\;(x^2)^{\frac{1}{3}}\cdot y^{\frac{1}{3}}\cdot x^{\frac{1}{2}}\cdot \left(y^3\right)^{\frac{1}{2}} = \;x^{\frac{2}{3}}\cdot y^{\frac{1}{3}}\cdot x^{\frac{1}{2}}\cdot y^{\frac{3}{2}}

    . . = \;x^{\frac{2}{3}}\cdot x^{\frac{1}{2}}\cdot y^{\frac{1}{3}}\cdot y^{\frac{3}{2}} \;=\;x^{(\frac{2}{3}+\frac{1}{2})}\cdot y^{(\frac{1}{3}+\frac{3}{2})}\;= \;\boxed{x^{\frac{7}{6}}y^{\frac{11}{6}}}



    7)\;\; 2\sqrt[3]{16}\cdot\sqrt[6]{4}

    We have: . 2\cdot(16)^{\frac{1}{3}}\cdot(4)^{\frac{1}{6}}\;=\  ;2\cdot(2^4)^{\frac{1}{3}}\cdot(2^2)^{\frac{1}{6}}  \;=\;2\cdot2^{\frac{4}{3}}\cdot2^{\frac{1}{3}}

    . . = \;2^{(1+\frac{4}{3}+\frac{1}{3})} \;=\;\boxed{2^{\frac{8}{3}}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    I'm not sure what final form the answer should take.

    For example, #6 is: . x^{\frac{7}{6}}y^{\frac{11}{6}} \:=\:\left(x^7y^{11}\right)^{\frac{1}{6}}\:=\:\sqr  t[6]{x^7y^{11}}

    And #7 is: . 2^{\frac{8}{3}}\:=\:2^{(2 + \frac{2}{3})} \:=\:2^2\cdot2^{\frac{2}{3}} \;=\;4\sqrt[3]{4}

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