Results 1 to 3 of 3

Math Help - Index Question

  1. #1
    Newbie
    Joined
    Jul 2010
    From
    Arbroath, Bonnie Scotland.
    Posts
    11

    Index Question

    Can someone check my calculation please?

    Question: (8x)^1/3 x x^3/4

    My working: (8x)^1/3 x x^3/4

    2x^1/3 x x^3/4

    = 2x^13/12

    Grateful for your comments.

    Cromlix
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Looks correct to me

    Just a note, when doing maths in plain text is is better to use * for multiply rather than x because x can get confused with the variable x

    For example 4*2 = 8
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,512
    Thanks
    1404
    This is nearly impossible to read because you're using x to represent both x and \times.

    I assume what you are writing is

    (8x)^{\frac{1}{3}}\times x^{\frac{3}{4}}

     = 8^{\frac{1}{3}} \times x^{\frac{1}{3}} \times x^{\frac{3}{4}}

     = 2x^{\frac{1}{3} + \frac{3}{4}}

     = 2x^{\frac{4}{12} + \frac{9}{12}}

     = 2x^{\frac{13}{12}}.


    So I agree with your final answer, but please learn some LaTeX to avoid confusing the people who are trying to help you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] a question about normal subgroups and index.
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: August 1st 2011, 08:37 PM
  2. Index Law
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 3rd 2011, 06:48 AM
  3. Replies: 2
    Last Post: November 13th 2008, 12:35 AM
  4. index
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: October 11th 2008, 03:36 PM
  5. 10-sec vol index
    Posted in the Statistics Forum
    Replies: 0
    Last Post: June 16th 2008, 01:47 PM

Search Tags


/mathhelpforum @mathhelpforum