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Math Help - Some confusion over some indices questions

  1. #1
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    Jun 2009
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    Some confusion over some indices questions

    I am currently doing some algebra practice for English pre A Level Maths. I am doing examples of a worksheet and I am finding a few problems that have cropped up so far that I would greatly appreciate if someone could explain how to get the right solution as my algebra is pretty patchy in places and needs to be mastered!

    Here are they are:

    Simplify the following expressions into their simplest form

    1,

     {\frac {ab^2}{cd}} \times {\frac {ac^2}{bd}}

    I simplfied to:

    ={\frac {a^2bc}{d^2}}

    The answer in the back of the book:

    = a^2bcd^2 ?

    2,

    {[(ab)^3]^{-\frac {1}{2}}}

    I simplified to:

    =[a^3b^3]^{-\frac {3}{2}}

    Answer in the back of the book:

    =a^{-\frac {1}{2}}b^{-\frac {3}{2}} ?

    3,

    (abc)^{\frac{1}{2}} \div (a^3b^3c^3)^{\frac {2}{3}}

    I simplified to:

     <br />
={\frac {a^{\frac{1}{2}}b^{\frac{1}{2}}c^{\frac{1}{2}}} {a^2b^2c^2}}<br />

    The answer im the back of the book:

     <br />
= a^{-\frac{3}{2}} b^{-\frac{3}{2}} c^{-\frac{3}{2}}<br />
?

    Thanks in advance.

    David
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  2. #2
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
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    1,261
    Quote Originally Posted by Duckers View Post
    I am currently doing some algebra practice for English pre A Level Maths. I am doing examples of a worksheet and I am finding a few problems that have cropped up so far that I would greatly appreciate if someone could explain how to get the right solution as my algebra is pretty patchy in places and needs to be mastered!

    Here are they are:

    Simplify the following expressions into their simplest form

    1,

     {\frac {ab^2}{cd}} \times {\frac {ac^2}{bd}}

    I simplfied to:

    ={\frac {a^2bc}{d^2}}

    The answer in the back of the book:

    = a^2bcd^2 ?

    2,

    {[(ab)^3]^{-\frac {1}{2}}}

    I simplified to:

    =[a^3b^3]^{-\frac {3}{2}}

    Answer in the back of the book:

    =a^{-\frac {1}{2}}b^{-\frac {3}{2}} ?

    3,

    (abc)^{\frac{1}{2}} \div (a^3b^3c^3)^{\frac {2}{3}}

    I simplified to:

     <br />
={\frac {a^{\frac{1}{2}}b^{\frac{1}{2}}c^{\frac{1}{2}}} {a^2b^2c^2}}<br />

    The answer im the back of the book:

     <br />
= a^{-\frac{3}{2}} b^{-\frac{3}{2}} c^{-\frac{3}{2}}<br />
?

    Thanks in advance.

    David

    (1) Your ans is correct .

    (2)  a^{-\frac{3}{2}}b^{-\frac{3}{2}}


    (3) a^{\frac{1}{2}-2}b^{\frac{1}{2}-2}c^{\frac{1}{2}-2}
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    (1) Your ans is correct .

    (2)  a^{-\frac{3}{2}}b^{-\frac{3}{2}}


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

    Thank you for confirming the answers, as I am unsure of how you got the last two answers would you mind shown how you solved them?

    Thanks again

    David
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  4. #4
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
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    1,261
    Quote Originally Posted by Duckers View Post
    Thank you for confirming the answers, as I am unsure of how you got the last two answers would you mind shown how you solved them?

    Thanks again

    David
    Ok sure .

    (2) [(ab)^3]^{-\frac{1}{2}}=(ab)^{-\frac{3}{2}}

     <br />
=a^{-\frac{3}{2}} b^{-\frac{3}{2}}<br />


    (3) division of indices -- ie \frac{a^m}{a^n}=a^{m-n}

    so follow my previous post , and simplify it .
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