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Math Help - simplifying expressions

  1. #1
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    simplifying expressions

    could someone help me with practice for my math placement in a few days. and sorry i didnt realize until after i posted that this is the elem/middle school section.

    Use the properties of exponents to simplify the expression


    could someone please show me how to get this answer?
    correct answer
    Last edited by rafaeli; June 29th 2008 at 09:59 AM.
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  2. #2
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    Quote Originally Posted by rafaeli View Post
    could someone help me with practice for my math placement in a few days.

    Use the properties of exponents to simplify the expression


    could someone please show me how to get this answer?
    correct answer
    \left(\frac{3a^{\frac12} b^{-\frac13}c} {a^{-\frac12}b^{\frac13}<br />
c^{\frac72}}\right)^{-3}

    As the bracket is to the power of -3, a minus means that that bracket is the reciprocal to 1.

    \frac{1}{\left(\frac{3a^{\frac12} b^{-\frac13}c} {a^{-\frac12}b^{\frac13} c^{\frac72}}\right)^3}

    This is equal to the bracket being flipped hence:

    \left(\frac{a^{-\frac12}b^{\frac13}<br />
c^{\frac72}}{3a^{\frac12} b^{-\frac13}c}\right)^{3}

    The general rule is (x^a)^b = x^{(a)(b)} = x^ {ab}

    \frac {a^{-\frac32} b<br />
c^{\frac{21}{2}}}{27 a^{\frac32} b^{-1} c^3}

    The general rule for simplifying fraction with variable is \frac{x^a}{x^b}  = x^{a-b} hence:

    \frac{a^{-3} b^2<br />
 c^{\frac{15}{2}}}{27}

    For indices which is in the form x^{-a}, this is equal to it being \frac{1}{x^a}, hence:

    \frac{b^2  c^{\frac{15}{2}}}{27a^3}
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  3. #3
    Senior Member nikhil's Avatar
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    Lightbulb Here it is

    Hi!rafaeli
    Recall that (a^n)/(b^m)=a^(n-m)
    so for a we have (a^(1/2))/(a^(-1/2))=a^((1/2)-(-1/2)=a^((1/2)+(1/2))=a
    similarly for b we have
    (b^(-1/3))/(b^(1/3))=b^(-2/3)
    similarly for c we have
    c/c^(7/2)=c^(-5/2)

    So the question can be written as T=[3ab^(-2/3)c^(-5/2)]^-3
    Now as x^(-n)=1/x^n and
    (x^a)^b=x^ab
    so T=3^(-3)a^(-3)b^2c^(15/2)
    which is =[b^2c^(15/2)]/[27a^3]
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by rafaeli View Post
    could someone help me with practice for my math placement in a few days. and sorry i didnt realize until after i posted that this is the elem/middle school section.

    Use the properties of exponents to simplify the expression


    could someone please show me how to get this answer?
    correct answer
    Edit: I know I'm late, but this is just another approach.

    First, let's make sue you know a couple of exponent rules:

    1. \frac{x^m}{x^n}=x^{m-n}

    2. (x^m)^n=x^{mn}

    3. \left(\frac{x}{y}\right)^{-m}=\left(\frac{y}{x}\right)^m

    Now,

    \left(\frac{3a^\frac{1}{2}b^\frac{-1}{3}c}{a^\frac{-1}{2}b^\frac{1}{3}c^\frac{7}{2}}\right)^{-3}

    Use rule 1 to get:

    \left(\frac{3a^1}{b^\frac{2}{3}c^\frac{5}{2}}\righ  t)^{-3}

    Use rule 3 to get:

    \left(\frac{b^\frac{2}{3}c^\frac{5}{2}}{3a}\right)  ^3

    Use rule 2 to finish:

    \frac{b^2 c^\frac{15}{2}}{27a^3}
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