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Math Help - [SOLVED] Need help with exponents: simplifying 1

  1. #1
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    [SOLVED] Need help with exponents: simplifying 1

    Hello, I was wondering if someone can tell me if I did these exponent problems right? I'm a bit confused.

    Convert the following to equivalent forms in which no negative exponents appear:

    a. (2/5)^-1 ---------- I said this equals -2/-5, which is 2/5

    b. 6/x^-2 (just the x is to the negative 2) ------ I said this was 6^2/x, which is 36/x

    c. (-3/2)^-3 --------- I said this was 81/-8, or -81/8

    d. 6xy/3x^-1y-2) (that looks weird anyway 6xy over 3 x to the negative one, y to the negative 2) ---- I was confused. I said 2x^2y^3 (2xsquaredycubed)

    e. (2x2/3x-1)^-2 -- I said this was -4x-4/-9x2 = 4x/9x^6.

    Note that x is the variable, not multiplication.

    Thanks!!

    Also, is x^3 times x^-3 x^0?

    Thank you so much
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  2. #2
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    Quote Originally Posted by view360 View Post
    Hello, I was wondering if someone can tell me if I did these exponent problems right? I'm a bit confused.

    Convert the following to equivalent forms in which no negative exponents appear:

    a. (2/5)^-1 ---------- I said this equals -2/-5, which is 2/5
    that is equals to \frac{5}{2}

    b. 6/x^-2 (just the x is to the negative 2) ------ I said this was 6^2/x, which is 36/x
    =\frac{x^2}{6}

    c. (-3/2)^-3 --------- I said this was 81/-8, or -81/8
    =-\frac{8}{27}

    d. 6xy/3x^-1y-2) (that looks weird anyway 6xy over 3 x to the negative one, y to the negative 2) ---- I was confused. I said 2x^2y^3 (2xsquaredycubed)
    is this the question? \frac{6xy}{3x^{-1}y^{-2}}

    =\frac{6x^2y^3}{3}

    =2x^2y^3

    note: there should be no negative exponents in the denominator


    e. (2x2/3x-1)^-2 -- I said this was -4x-4/-9x2 = 4x/9x^6.
    (\frac{2x^2}{3x-1})^{-2}

    \frac{9x^2-6x+1}{4x^4}

    Also, is x^3 times x^-3 x^0?
    x^3 * x^-3 * x^0

    x^3 * \frac{1}{x^3} * 1

    =1
    Last edited by princess_21; February 3rd 2009 at 06:29 PM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Some minor mistakes here
    Quote Originally Posted by princess_21 View Post
    =\frac{x^2}{6}
    \frac 6{x^{-2}} = 6x^2

    =\frac{8}{27}
    this should be -8/27, you left off the negative sign

    also, when superscripts or subscripts have more than one term, you need to put them in { } brackets for the LaTeX to look nice.


    [tex]x^-2y^2[/tex] \longrightarrow x^-2y^2

    [tex]x^{-2}y^2[/tex] \longrightarrow x^{-2}y^2


    see, we have -2 as the power, that's two things, a minus sign and a 2, were it just a 2, typing x^2 is fine, but for the -2 we need to enclose it x^{-2}
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  4. #4
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    Thanks, Jhevon!!
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    also, this is not right. we can distribute powers across products, not sums

    (a + b)^2 \ne a^2 + b^2 for instance, but rather (a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2

    similarly, (x^{-1} + y^{-1})^{-2} = \frac 1{(x^{-1} + y^{-1})^2} = \frac 1{x^{-2} + 2(xy)^{-1} + y^{-2}}


    if you don't like negative powers in the denominator, you could approach it this way

    (x^{-1} + y^{-1})^{-2} = \left( \frac 1x + \frac 1y \right)^{-2} = \left( \frac {x + y}{xy} \right)^{-2} = \frac {x^2y^2}{(x + y)^2} = \frac {x^2y^2}{x^2 + 2xy + y^2}



    also, please post new problems in a new thread. it saves me the work of having to split the thread up
    sorry my mistake.
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  6. #6
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    Thanks to both of you, thank you so much for taking the time to help me out. I think I'm finally getting the hang of these now!!
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