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Math Help - Algebra 2 help needed on exponents and logs

  1. #1
    Newbie JoanneMac's Avatar
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    Exclamation Algebra 2 help needed on exponents and logs

    OK, I'm baaack!

    Need help on the following problems and they need to be simplified:

    1. -x^2y^2/x^1/2y^0=?

    2. cubed root of "a" multplied by fifth root of "a"=??

    And is this right???: cubed root of X^5y^5= (x^5y^5)^1/3=x^5/3y^5/3=(xy)^5/3

    Thanks for your excellent help
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  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
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    Quote Originally Posted by JoanneMac View Post
    OK, I'm baaack!

    Need help on the following problems and they need to be simplified:

    1. -x^2y^2/x^1/2y^0=?

    2. cubed root of "a" multplied by fifth root of "a"=??


    Hello Joanne,

    I'm not sure if I translated your first one correctly. You should've used parentheses to group things.

    -\dfrac{x^2y^2}{x^{\frac{1}{2}}y^0}=-x^{\frac{3}{2}}y^2

    \sqrt[3]{a} \cdot \sqrt[5]{a}=a^{\frac{1}{3}}a^{\frac{1}{5}}=a^{\frac{5}{15}  +\frac{3}{15}}=a^{\frac{8}{15}}

    Quote Originally Posted by JoanneMac View Post
    And is this right???: cubed root of X^5y^5= (x^5y^5)^1/3=x^5/3y^5/3=(xy)^5/3

    Thanks for your excellent help
    Looks good.
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  3. #3
    Member
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    Indiana
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    Quote Originally Posted by JoanneMac View Post
    OK, I'm baaack!

    Need help on the following problems and they need to be simplified:

    1. -x^2y^2/x^1/2y^0=?

    2. cubed root of "a" multplied by fifth root of "a"=??

    And is this right???: cubed root of X^5y^5= (x^5y^5)^1/3=x^5/3y^5/3=(xy)^5/3

    Thanks for your excellent help
    Hi JoanneMac,

    It is hard to tell what was the original problem. Are you sure all your parentheses are where they are supposed to?
    In other word, do you really mean -x^2y^2/x^1/2y^0? which is  \frac{(-x^2)*(y^2)}{(x^1)*(2y^0)}.
    Or do you mean (-x^(2y^2))/(x^(1/2y^0))? Which is  \frac{-x^(2y^2)}{x^(\frac{1}{2}y^0)}.
    Or something else?
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  4. #4
    Newbie JoanneMac's Avatar
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    Red face The poster above..........

    Masters got the equations right. Sorry, I should've used parentheses.
    Thanks for your help, though.
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