Results 1 to 6 of 6

Thread: help with the powers on my first post

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    3

    help with the powers on my first post

    Reduce to the lowest power?

    $\displaystyle /sqrt^6/frac{3}-{8}/frac/frac{1}-{4}$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    681
    Thanks
    1
    Quote Originally Posted by gillyman View Post
    Reduce to the lowest power?

    $\displaystyle /sqrt^6/frac{3}-{8}/frac/frac{1}-{4}$
    could you rewrite this using latex
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2009
    Posts
    3
    Hope this is correct, its also my first time using LATEX
    $\displaystyle (/sqrt)^{6}-/frac{3}{8}/frac{1}{4}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    681
    Thanks
    1

    Cool

    Quote Originally Posted by gillyman View Post
    Hope this is correct, its also my first time using LATEX
    $\displaystyle (/sqrt)^{6}-/frac{3}{8}/frac{1}{4}$

    $\displaystyle \sqrt{6}-\frac{\frac{3}{8}}{\frac{1}{4}}$

    or

    $\displaystyle \frac{\sqrt{6}}{\frac{3}{8}} - \frac{8}{\frac{1}{4}}$

    sorry i know this is a pain but...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2009
    Posts
    3

    Its neither of these, Sorry

    $\displaystyle \frac{(\sqrt)^6\frac{3}{8}}{\frac{1}{4}}$

    I hope this is correct. I do need a crash course in LATEK.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,798
    Thanks
    3035
    Do you mean $\displaystyle \frac{\sqrt[6]{\frac{3}{8}}}{\frac{1}{4}}$?

    That is $\displaystyle 4(3^{1/6})2^{1/2})= 2^{5/2}3^{1/6}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Not sure exactly where to post this
    Posted in the Algebra Forum
    Replies: 8
    Last Post: Jan 17th 2011, 04:49 PM
  2. My first post here
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Apr 13th 2010, 01:30 PM
  3. First post
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Dec 8th 2009, 12:50 PM

Search Tags


/mathhelpforum @mathhelpforum