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Math Help - Surds and fractional indicies

  1. #1
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    Surds and fractional indicies

    I'm starting with basic simplification and while I thought I understood the rules I cannot seem to get the same answers as the book I am learning from.

    The two questions that are vexing me are as follows:

    Simplify:

    1. (3^3)^(1/2) * 9^(1/4)


    2.  CubeRoot(a^4b^5) * (b^(1/3))/a

    The answer for the first is 3^2 but I get 27^(7/4).

    The second really makes me doubt by understanding as I swapped the first part around to read:

    a^(4/3)b^(5/3) yet the book states this should be a^(3/4)b^(5/3).

    Can anyone see where I am going wrong?
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  2. #2
    Super Member craig's Avatar
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    For the first one.

    (3^3)^{\frac{1}{2}} * 9^(1/4)
    =
    3^{\frac{3}{2}} * (3^2)^{\frac{1}{4}}
    =
    3^{\frac{3}{2}} * 3^{\frac{1}{2}}

    which equals?

    Read through this example looking at what I have done and attempt the second one again.

    If you do not get it right post your working and we can go through where you went wrong.
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  3. #3
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    Thanks for the help Craig, answering the first has let me answer most of the other questions but I still seem to be drawing a blank on the second.

    So the question is:

    \sqrt[3]{a^4b^5}*\sqrt[3]b/a

    which could be written as:

     (a^4)^\frac{1}{3}(b^5)^\frac{1}{3}*b^\frac{1}{3}*a  ^{-1})

    So

     (a^\frac{4}{3})*(b^\frac{5}{3})*(b^\frac{1}{3})*(a  ^{-1})

    To get the above answer would mean that:



    Using the multiplication rule on the powers of a and b respectively gives:

     (a^\frac{1}{3})*(b^\frac{6}{3})

    or

     (a^\frac{1}{3})*(b^2)

    However the answer in the book is:

     (a^\frac{-1}{4})(b^2)

    To get this answer the first simplification would need to be:

     (a^\frac{3}{4})(b^\frac{5}{3})*b^\frac{1}{3}*a^{-1})

    but I have no idea how you would get this result for the first a...
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by AliceFisher View Post
    Thanks for the help Craig, answering the first has let me answer most of the other questions but I still seem to be drawing a blank on the second.

    So the question is:

    \sqrt[3]{a^4b^5}*\sqrt[3]b/a

    which could be written as:

     (a^4)^\frac{1}{3}(b^5)^\frac{1}{3}*b^\frac{1}{3}*a  ^{-1})

    So

     (a^\frac{4}{3})*(b^\frac{5}{3})*(b^\frac{1}{3})*(a  ^{-1})

    To get the above answer would mean that:



    Using the multiplication rule on the powers of a and b respectively gives:

     (a^\frac{1}{3})*(b^\frac{6}{3})

    or

     (a^\frac{1}{3})*(b^2)

    However the answer in the book is:

     (a^\frac{-1}{4})(b^2)

    To get this answer the first simplification would need to be:

     (a^\frac{3}{4})(b^\frac{5}{3})*b^\frac{1}{3}*a^{-1})

    but I have no idea how you would get this result for the first a...
    As far as the question you have posted goes (if your question is \sqrt[3]{a^4b^5}*\frac{\sqrt[3]b}{a}),

    the answer you have got , that is,  (a^\frac{1}{3})*(b^2)
    looks correct.

    If the answer given by your book is  (a^\frac{-1}{4})(b^2), the question should look somewhat like this: \sqrt[3]{a^4b^5}*\frac{\sqrt[3]b}{\sqrt[12]{a^{19}}}
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  5. #5
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    Well that would certainly explain why I have had so many problems trying to reconcile my answer with the books!

    Thanks hashish!
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by AliceFisher View Post
    Well that would certainly explain why I have had so many problems trying to reconcile my answer with the books!

    Thanks hashish!
    AliceFisher

    Its Harish, not hashish!

    hahah
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  7. #7
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    Oh dear
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