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  1. #1
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    simplify

    1/ $\displaystyle 3\sqrt{(x^3y^4)^2}$

    2/ $\displaystyle \frac{(x^2y^4)^2}{x^5y^-1}$

    3/ (x^3a+1)(x^a-3)/4sqrtx^8
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  2. #2
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    Quote Originally Posted by william View Post

    1/ $\displaystyle 3\sqrt{(x^3y^4)^2}$

    2/ $\displaystyle \frac{(x^2y^4)^2}{x^5y^-1}$

    3/ $\displaystyle \frac{(x^{3a}+1)(x^a-3)}{4\sqrt{x^8}}$
    Hello william,

    [1] $\displaystyle 3\sqrt{(x^3y^4)^2}=3x^3y^4$

    [2] $\displaystyle \frac{(x^2y^4)^2}{x^5y^-1}=\frac{x^4y^8y}{x^5}=\frac{x^9}{x}$

    [3] $\displaystyle \frac{(x^{3a}+1)(x^a-3)}{4\sqrt{x^8}}=\frac{x^{4a}-3x^{3a}+x^{a}-3}{4x^4}$

    I made a few corrections on [3]. I'm not sure I translated your 3rd one correctly. Let me know.
    Last edited by masters; Jan 14th 2009 at 01:02 PM.
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  3. #3
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    Quote Originally Posted by masters View Post
    Hello william,

    [1] $\displaystyle \frac{(x^2y^4)^2}{x^5y^-1}=3x^3y^4$

    [2] $\displaystyle \frac{(x^2y^4)^2}{x^5y^-1}=\frac{x^4y^8y}{x^5}=\frac{x^9}{x}$

    [3] $\displaystyle \frac{(x^{3a+1})(x^{a-3}}{4\sqrt{x^8}}$

    I made a few corrections on [3]. I'm not sure I translated your 3rd one correctly. Let me know.
    Thanks alot

    What did you do there for 1, I am assuming you just wrote the second question as the first by mistake. For 3,

    For 2 how did you get y^8 and not y^16, can you show steps please?

    $\displaystyle \frac{(x^{3a+1})(x^{a-3}}{4\sqrt{x^8}}$

    in the numerator it is x^{3a+1} and x^{a-3}

    and if you don't mind me asking how do i simplify (27)^-2/3
    Last edited by william; Jan 14th 2009 at 12:57 PM.
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  4. #4
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    Quote Originally Posted by william View Post
    Thanks alot

    What did you do there for 1, I am assuming you just wrote the second question as the first by mistake. For 3,

    $\displaystyle \frac{(x^{3a+1})(x^{a-3}}{4\sqrt{x^8}}$

    in the numerator it is x^{3a+1} and x^{a-3}

    and if you don't mind me asking how do i simplify (27)^-2/3
    William,

    I don't have a clue what happened there. Some things did get out of whack, didn't they. Maybe you and I were editing at the same time and maybe divided by zero and clobbered the database.

    [1] $\displaystyle 3\sqrt{(x^3y^4)^2}=3x^3y^4$
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  5. #5
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    Quote Originally Posted by masters View Post
    William,

    I don't have a clue what happened there. Some things did get out of whack, didn't they. Maybe you and I were editing at the same time and maybe divided by zero and clobbered the database.

    [1] $\displaystyle 3\sqrt{(x^3y^4)^2}=3x^3y^4$
    I appreciate your replies but I am not sure how you are arriving at these answers
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  6. #6
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    Quote Originally Posted by william View Post
    I appreciate your replies but I am not sure how you are arriving at these answers
    William,

    Do you know that $\displaystyle \sqrt{a^2}=a$?

    That's what I used to answer the first one. You could also look at it this way:

    $\displaystyle \sqrt{a}=\sqrt{a^1}=a^{1/2}$. You can convert this into rational exponents by dividing the power in the radicand by the index of the root.

    $\displaystyle 3\sqrt{(x^3y^4)^2}=3(x^3y^4)^{2/2}=3(x^3 y^4)^1=3x^3 y^4$

    Or, you could expand the radicand first and then simplify at the end like

    $\displaystyle 3 \sqrt{(x^3y^4)^2}=3\sqrt{x^6y^8}=3x^{6/2}y^{8/2}=3x^3y^4$
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  7. #7
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    Quote Originally Posted by masters View Post
    William,

    Do you know that $\displaystyle \sqrt{a^2}=a$?

    That's what I used to answer the first one. You could also look at it this way:

    $\displaystyle \sqrt{a}=\sqrt{a^1}=a^{1/2}$. You can convert this into rational exponents by dividing the power in the radicand by the index of the root.

    $\displaystyle 3\sqrt{(x^3y^4)^2}=3(x^3y^4)^{2/2}=3(x^3 y^4)^1=3x^3 y^4$

    Or, you could expand the radicand first and then simplify at the end like

    $\displaystyle 3 \sqrt{(x^3y^4)^2}=3\sqrt{x^6y^8}=3x^{6/2}y^{8/2}=3x^3y^4$
    you have been great help

    if i wanted to simplify (27)^{-2/3} would -8 be correct?
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  8. #8
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    Quote Originally Posted by william View Post
    Thanks alot

    What did you do there for 1, I am assuming you just wrote the second question as the first by mistake. For 3,

    For 2 how did you get y^8 and not y^16, can you show steps please?
    Use this property of exponents: $\displaystyle (a^m)^n=a^{mn}$. So $\displaystyle (y^4)^2=y^{4 \cdot 2}=y^8$

    Quote Originally Posted by william View Post

    $\displaystyle \frac{(x^{3a+1})(x^{a-3}}{4\sqrt{x^8}}$

    in the numerator it is x^{3a+1} and x^{a-3}
    Is it $\displaystyle \frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}$

    If so, $\displaystyle \frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}=\frac{x^{(3a+3)+(a-3)}}{4x^4}=\frac{x^{4a}}{4x^4}=\frac{x^{4a-4}}{4}=\frac{x^{4(a-1)}}{4}$

    Quote Originally Posted by william View Post
    and if you don't mind me asking how do i simplify (27)^-2/3
    $\displaystyle 27^{-\frac{2}{3}}=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{( \sqrt[3]{27})^2}=\frac{1}{(3)^2}=\frac{1}{9}$
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  9. #9
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    Quote Originally Posted by masters View Post
    Use this property of exponents: $\displaystyle (a^m)^n=a^{mn}$. So $\displaystyle (y^4)^2=y^{4 \cdot 2}=y^8$



    Is it $\displaystyle \frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}$

    If so, $\displaystyle \frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}=\frac{x^{(3a+3)+(a-3)}}{4x^4}=\frac{x^{4a}}{4x^4}=\frac{x^{4a-4}}{4}=\frac{x^{4(a-1)}}{4}$



    $\displaystyle 27^{-\frac{2}{3}}=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{( \sqrt[3]{27})^2}=\frac{1}{(3)^2}=\frac{1}{9}$
    Thank you very much for all the help you are a great mathematician and an excellent teacher. Cheers!
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