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Math Help - [SOLVED] SAT math page 674 #19

  1. #1
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    [SOLVED] SAT math page 674 #19

    If k, n, x and y are positive numbers satisfying x^\frac{-4}{3}=k^{-2} and y^\frac{4}{3}=n^2, what is (xy)^\frac{-2}{3} in terms of n and k?

    Thanks in advance!! (:
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  2. #2
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    this will get you started ...

    x^{-\frac{4}{3}} = k^{-2}

    \left(x^{-\frac{4}{3}}\right)^{-\frac{3}{4}} = \left(k^{-2}\right)^{-\frac{3}{4}}

    x = k^{\frac{3}{2}}

    now do the same to solve for y in terms of n ... then evaluate (xy)^{-\frac{2}{3}}
    Last edited by skeeter; September 28th 2008 at 07:03 AM. Reason: fix my "dyslexic" fraction
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  3. #3
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    Quote Originally Posted by skeeter View Post
    this will get you started ...



    \left(x^{-\frac{4}{3}}\right)^{-\frac{3}{4}} = \left(k^{-2}\right)^{-\frac{4}{3}}
    How did you get this? Is this a rule? why multiply x with ^(-3/4)?

    Thanks in advance!!
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  4. #4
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    Quote Originally Posted by fabxx View Post
    How did you get this? Is this a rule? why multiply x with ^(-3/4)?

    Thanks in advance!!
    (a^b)^c=(a^c)^b=a^{bc}

    ____________________________________________
    other rules of exponents (among others) :
    a^b a^c=a^{b+c}

    \frac{a^b}{a^c}=a^{b-c}
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  5. #5
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    Quote Originally Posted by fabxx View Post
    How did you get this? Is this a rule? why multiply x with ^(-3/4)?

    Thanks in advance!!
    my apologies ... I assumed you had some knowledge regarding the Laws of Exponents
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    Quote Originally Posted by fabxx View Post
    If k, n, x and y are positive numbers satisfying x^\frac{-4}{3}=k^{-2} and y^\frac{4}{3}=n^2, what is (xy)^\frac{-2}{3} in terms of n and k?

    Thanks in advance!! (:
    Basically,

    If its SATs, AS A RULE, forget about complex rules.

     x ^\frac{-4}{3} = k^{-2} ...(1)

    and

     y^\frac{4}{3} = n^2 ... (2)

    From (1),

     \frac{1}{x ^\frac{4}{3}} = \frac{1}{k^2}

    Cross multiply and shift the x-term to the left hand side, you get:

    x ^\frac{4}{3} = k^2 ... (3)

    Multiply (2) and (3)

    x ^\frac{4}{3} * y^\frac{4}{3} =  n^2* k^2

    (xy) ^\frac{4}{3} = (nk)^2

    ((xy)^\frac{2}{3})^2 = (nk)^2

    Squaring on both sides and making a reciprocal of the left-hand and the right-hand-side term gives:

    (xy)^\frac{-2}{3}=\frac{1}{nk} <= Answer
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  7. #7
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    i got x=k^(8/3) and y=n^(3/2)

    If i substitute I have (k^\frac{8}{3}n^\frac{3}{2} )^\frac{-2}{3}
    How do i solve after this?
    Like this? : k^\frac{-16}{9}n^{-1} but how do i get the final answer \frac{1}{nk}
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  8. #8
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    had to fix my original post ... had the fraction flipped.

    should be x = k^{\frac{3}{2}}

    so, you should get ...

    (k^{\frac{3}{2}}n^{\frac{3}{2}})^{-\frac{2}{3}} = k^{-1}n^{-1} = \frac{1}{kn}
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  9. #9
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    Find shortcuts - do not solve for x. See my steps how I got the answer.

    Hope that helps.
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  10. #10
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    Thanks to all three of you, I know how to do it now. Thanks guys. You guys were very helpful!!
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