Results 1 to 9 of 9

Math Help - simplify

  1. #1
    Senior Member
    Joined
    Oct 2008
    Posts
    323

    simplify

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

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

    3/ (x^3a+1)(x^a-3)/4sqrtx^8
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by william View Post

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

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

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

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

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

    [3] \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; January 14th 2009 at 01:02 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Oct 2008
    Posts
    323
    Quote Originally Posted by masters View Post
    Hello william,

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

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

    [3] \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?

    \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; January 14th 2009 at 12:57 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    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,

    \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] 3\sqrt{(x^3y^4)^2}=3x^3y^4
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Oct 2008
    Posts
    323
    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] 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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    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 \sqrt{a^2}=a?

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

    \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.

    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

    3 \sqrt{(x^3y^4)^2}=3\sqrt{x^6y^8}=3x^{6/2}y^{8/2}=3x^3y^4
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Oct 2008
    Posts
    323
    Quote Originally Posted by masters View Post
    William,

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

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

    \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.

    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

    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    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: (a^m)^n=a^{mn}. So (y^4)^2=y^{4 \cdot 2}=y^8

    Quote Originally Posted by william View Post

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

    If so, \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
    27^{-\frac{2}{3}}=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{(  \sqrt[3]{27})^2}=\frac{1}{(3)^2}=\frac{1}{9}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Oct 2008
    Posts
    323
    Quote Originally Posted by masters View Post
    Use this property of exponents: (a^m)^n=a^{mn}. So (y^4)^2=y^{4 \cdot 2}=y^8



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

    If so, \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}



    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. simplify #2
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 21st 2009, 02:37 PM
  2. Please simplify
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 27th 2009, 04:32 AM
  3. Can you simplify this further?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 23rd 2009, 06:45 AM
  4. Simplify
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 12th 2009, 05:06 AM

Search Tags


/mathhelpforum @mathhelpforum