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Math Help - Exponents power to a power

  1. #1
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    Exponents power to a power

    Wondering if someone can check my work, thanks!

    (5x^3y^2)^3
    I got 125x^9y^6

    (3ab^5)^2(a^4b^6)
    I got 9a^6b^16

    (-2fg^2h^7)^4
    I got -16f^4g^8h^28

    (3xy)^3(4x^5y^2)^2
    I got 432x^13y^4

    Also was wondering if anyone can help me with this probelm
    -(4x)^3(2x)^2(x^5)^5
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  2. #2
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    Quote Originally Posted by Jubbly View Post
    Wondering if someone can check my work, thanks!

    (5x^3y^2)^3
    I got 125x^9y^6
    correct

    (3ab^5)^2(a^4b^6)
    I got 9a^6b^16
    correct

    (-2fg^2h^7)^4
    I got -16f^4g^8h^28
    (-1)<sup>4</sup>= 1 not -1.

    (3xy)^3(4x^5y^2)^2
    I got 432x^13y^4
    You forgot the first y^3

    Also was wondering if anyone can help me with this probelm
    -(4x)^3(2x)^2(x^5)^5
    4^3= 64, 2^2= 4 and 64(4)= 256. (x^3)(x^2)(x^5)= x^{3+ 2+ 5}= x^{10}. And notice that the "-" is outside any power.
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  3. #3
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    Uhm.. for this equation (-2fg^2h^7)^4
    I don't understand you, other than that thanks for helping!
    Also for the last one since the negative is outside will the numbers still be negative?
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Jubbly View Post
    Uhm.. for this equation (-2fg^2h^7)^4
    I don't understand you, other than that thanks for helping!
    (-1)^4 = 1 .....sign is inside bracket else you are correct

    Also for the last one since the negative is outside will the numbers still be negative?
    Yes you are correct
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