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Math Help - Simplify using Exponent Laws

  1. #1
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    Simplify using Exponent Laws

    I have two questions, if any help is given I'd appreciate it.

    Simplify using the laws of exponents.

    1) (3t - 2t^(-1)) / t^3

    2) (3p^2 - p^(-3)) / p^4

    Thank you very much.
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    Quote Originally Posted by Jeavus View Post
    I have two questions, if any help is given I'd appreciate it.

    Simplify using the laws of exponents.

    1) (3t - 2t^(-1)) / t^3
    \frac {3t - 2t^{-1}}{t^3} = \frac {3t}{t^3} - \frac {2t^{-1}}{t^3}

    = 3 t^{1 - 3} - 2t^{-1-3}

    Now simplify that.

    The second question is done similarly
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  3. #3
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    Can you explain how you got:

    3 t^{1 - 3} - 2t^{-1-3}

    Thanks.

    EDIT: Nevermind, sorry.
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    Quote Originally Posted by Jeavus View Post
    Can you explain how you got:

    3 t^{1 - 3} - 2t^{-1-3}


    Thanks.
    There is a law of exponents that says \frac {x^a}{x^b} = x^{a - b}

    so, for instance, \frac {3t}{t^3} = 3 \cdot \frac {t}{t^3} = 3 t^{1 - 3}, since t = t^1
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  5. #5
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    So my final answers were:

    (3t - 2t^(-1)) / t^3
    = 3t / t^3 - 2t^(-1)/t^3
    = 3t^(1-3) - 2t^(-1-3)
    = 3t^(-2) - 2t^(-4)
    = (3/t^2) - (2/t^4)
    = (3t^2 - 2)/t^4

    Is that correct?

    (3p^2 - p^(-3))/p^4
    = 3p^(2-4) - p^(-3-4)
    = 3p^(-2) - p^(-7)
    = (3/p^2) - (1/p^7)
    = (3p^5 - 1) / p^7

    Is that correct?
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  6. #6
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    Quote Originally Posted by Jeavus View Post
    So my final answers were:

    (3t - 2t^(-1)) / t^3
    = 3t / t^3 - 2t^(-1)/t^3
    = 3t^(1-3) - 2t^(-1-3)
    = 3t^(-2) - 2t^(-4)
    = (3/t^2) - (2/t^4)
    = (3t^2 - 2)/t^4

    Is that correct?

    (3p^2 - p^(-3))/p^4
    = 3p^(2-4) - p^(-3-4)
    = 3p^(-2) - p^(-7)
    = (3/p^2) - (1/p^7)
    = (3p^5 - 1) / p^7

    Is that correct?
    those don't really look simpler to me. you eliminated some of the variables, but the ones that remain have even higher powers than before. i would leave them separated as sums
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  7. #7
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    How do I go about solving a problem like this?

    1) (4 - (square root)x)/x^(3/2)
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  8. #8
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    Quote Originally Posted by Jeavus View Post
    How do I go about solving a problem like this?

    1) (4 - (square root)x)/x^(3/2)
    you do this the same as the last two, except you must know that \sqrt {x} = x^{\frac {1}{2}}

    see post #3 here to learn how to change various root-forms into powers.
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  9. #9
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    And do you know how to go abouts solving these two? Last two I promise.

    1) (x-9) / (x^(1/2)-3)

    2) (x-1)/((square root)x - x)
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    Quote Originally Posted by Jeavus View Post
    And do you know how to go abouts solving these two? Last two I promise.

    1) (x-9) / (x^(1/2)-3)
    Notice that the top is the difference of two squares in terms of \sqrt {x}

    \frac {x - 9}{x^{1/2} - 3} = \frac {\left( \sqrt {x} \right)^2 - 3^2}{\sqrt {x} - 3}

    = \frac { \left( \sqrt {x} + 3 \right) \left( \sqrt {x} - 3 \right)}{\sqrt {x} - 3}

    Hopefully the last line is obvious to you


    2) (x-1)/((square root)x - x)
    Note that \frac {x - 1}{ \sqrt {x} - x} = \frac {x - 1}{x \left( x^{-1/2} - 1 \right)}

    how do you think you should proceed?

    Hint: You do the same trick with the top as I did in the last question
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  11. #11
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    Thanks so much for all your help.

    PS: I did what you requested! :P
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  12. #12
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    Quote Originally Posted by Jeavus View Post
    Thanks so much for all your help.

    PS: I did what you requested! :P
    ok, just note that we have x^{{\color {red} -} 1/2} = \frac {1}{x^{1/2}} in the denominator, not x^{1/2}, so you can't just cancel. more simplification is needed before you cancel anything
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