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Quick maths question

  1. Nov 6, 2004 #1
    Hello,

    please can someone help me with this :

    Is 1 / (ab/d)^x/y the same as (d/ab) ^x/y ?

    Does the order matter ?



    Also, if I have to simplify root 10/root 160 and put it into surd form is 1/4
    wrong ?


    thanks


    Roger
     
  2. jcsd
  3. Nov 6, 2004 #2
    [tex]\frac{1}{(ab/d)^{x/y}}=(\frac{1}{(ab/d)})^{x/y}=(\frac{d}{ab})^{x/y}[/tex]

    So yes, it is the same

    Also:

    [tex]\frac{\sqrt{10}}{\sqrt{160}}=\sqrt{\frac{10}{160}}=\sqrt{\frac{1}{16}}=\frac{1}{4}[/tex]

    So you are again correct
     
    Last edited: Nov 6, 2004
  4. Nov 6, 2004 #3

    arildno

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    I assume you are talking about the expression:
    [tex]\frac{1}{(\frac{ab}{d})^{\frac{x}{y}}}[/tex]
    By ordinary rules of arithmetic, we have:
    [tex]\frac{1}{(\frac{ab}{d})^{\frac{x}{y}}}=\frac{(1)^{\frac{x}{y}}}{(\frac{ab}{d})^{\frac{x}{y}}}=[/tex]
    [tex](\frac{1}{\frac{ab}{d}})^{\frac{x}{y}}=(\frac{d}{ab})^{\frac{x}{y}}[/tex]

    Secondly, we have for positive, real numbers a,b:
    [tex]\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}[/tex]
    EDIT:
    Hmm..dawillem beat me here..
     
  5. Nov 6, 2004 #4
    Dear Arildno,

    The bit that says by the ordinary rules of arithmetic we have....

    Why did you apply the x/y to the top and bottom ?
    I thought it only applies to whats inside the brackets at the bottom ?
    Please can you explain this for me



    Also for the last bit, on roots, if it was root minus x / root minus y what is the general answer for that ?

    Thanks


    Roger
     
  6. Nov 6, 2004 #5

    Zurtex

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    It just made it easier and 1a=1 for any value of a.
     
  7. Nov 6, 2004 #6
    Response to your first question: My understanding is the expressions are equivalent, as would be
    1/((ab^x/y)/d^x/y) and (d^x/y)/(ab^xy). You need to observe any hierarchy is all.

    The second is numerically correct. Unless you need to express it as root 1/16 or 1/root 16 for some reason.
     
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