Results 1 to 10 of 10

Math Help - [SOLVED] Indice Operations Simplification Problem

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    6

    [SOLVED] Indice Operations Simplification Problem

    okay, so here's the problem. I wish I had tutoring when I was younger or had more help in math, now I'm trying to cram algebra revision, trig and curves and functions in 9 days for a test so I can get into engineering related classes I need.



    I get most problems, just here and there I stumble onto something I can't solve because I don't know the little points.


    So this is the equation:

    a^3 b^1/2 c^-1/2 (ab)^1/3
    ------------------------------ the "divided by" line

    ( squareroot(a^3 squareroot(b c)


    ok, i'm trying to figure out if the (ab) portion becomes a^1/3 and b^1/3

    and if the squareroot of a^3 is actually a^3/2 ,


    I'm pretty confused.



    p.s. I'm working from the Higher Engineering Mathematics Textbook (this problem specifically) and College Algebra Demystified and reviewing Algebra Demystified.



    and the answer is also there in the textbook, but it doesn't show you how to get it, and I've been trying to work the problem backwards too.

    Thanks.
    Last edited by immengin101; December 16th 2009 at 06:24 AM. Reason: more info
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by immengin101 View Post
    ...
    So this is the equation:

    a^3 b^1/2 c^-1/2 (ab)^1/3
    ------------------------------ the "divided by" line

    ( squareroot(a^3 squareroot(b c)


    ok, i'm trying to figure out if the (ab) portion becomes a^1/3 and b^1/3

    and if the squareroot of a^3 is actually a^3/2 ,

    ...
    If I understand your question correctly you want to simplify

    \dfrac{a^3 b^{\frac12} c^{-\frac12} \cdot (a b)^{\frac13}}{\sqrt{a^3 \cdot \sqrt{b c}}}

    Transform into a product:

    a^3 \cdot b^{\frac12} \cdot c^{-\frac12} \cdot a^{\frac13} \cdot b^{\frac13} \cdot a^{-\frac32} \cdot b^{-\frac14} \cdot c^{-\frac14}

    Collect like terms:

    a^{3+\frac13 - \frac32} \cdot b^{\frac12+\frac13-\frac14} \cdot c^{-\frac12-\frac34}

    I'll leave the rest for you.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Stroodle's Avatar
    Joined
    Jun 2009
    Posts
    367
    Quote Originally Posted by immengin101 View Post
    ok, i'm trying to figure out if the (ab) portion becomes a^1/3 and b^1/3

    and if the squareroot of a^3 is actually a^3/2 ,

    Hi there.

    (ab)^{\frac{1}{3}}=a^{\frac{1}{3}}\times b^{\frac{1}{3}}

    and

    \sqrt{a^3}=(a^3)^{\frac{1}{2}}=a^{\frac{3}{2}}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2009
    Posts
    6
    the
    "c" isn't squarerooted under the divided by line

    it looks like this

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2009
    Posts
    6
    Quote Originally Posted by Stroodle View Post
    Hi there.

    (ab)^{\frac{1}{3}}=a^{\frac{1}{3}}\times b^{\frac{1}{3}}

    and

    \sqrt{a^3}=(a^3)^{\frac{1}{2}}=a^{\frac{3}{2}}
    thank you.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68
    Quote Originally Posted by immengin101 View Post
    a^3 b^1/2 c^-1/2 (ab)^1/3
    ------------------------------ the "divided by" line
    ( squareroot(a^3) squareroot(b c)
    ok, i'm trying to figure out if the (ab) portion becomes a^1/3 and b^1/3
    and if the squareroot of a^3 is actually a^3/2 ,
    To start, more brackets required, and squareroot is same as ^(1/2); so:

    a^3 b^(1/2) c^(-1/2) (ab)^(1/3)
    --------------------------------- the "divided by" line
    (a^3)^(1/2) (bc)^(1/2)

    On your 2 "trying to figure out":
    (ab)^(1/3) = a^(1/3) b^(1/3)
    sqrt(a^3) = (a^3)^(1/2) = a^(3/2)
    Looks like you know what you're doing!

    Next step is separate (ab) and (bc):
    a^3 b^(1/2) c^(-1/2) a^(1/3) b^(1/3)
    -------------------------------------- the "divided by" line
    a^(3/2) b^(1/2) c^(1/2)

    Now, using rules:
    a^x a^y = a(x+y)
    a^x / a^y = a^(x-y) or 1/a^(y-z)
    you should be able to wrap that up in this neat little xmas package:
    a^(11/6) b^(1/3) / c
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2009
    Posts
    6
    Quote Originally Posted by Wilmer View Post
    To start, more brackets required, and squareroot is same as ^(1/2); so:

    a^3 b^(1/2) c^(-1/2) (ab)^(1/3)
    --------------------------------- the "divided by" line
    (a^3)^(1/2) (bc)^(1/2)

    On your 2 "trying to figure out":
    (ab)^(1/3) = a^(1/3) b^(1/3)
    sqrt(a^3) = (a^3)^(1/2) = a^(3/2)
    Looks like you know what you're doing!

    Next step is separate (ab) and (bc):
    a^3 b^(1/2) c^(-1/2) a^(1/3) b^(1/3)
    -------------------------------------- the "divided by" line
    a^(3/2) b^(1/2) c^(1/2)

    Now, using rules:
    a^x a^y = a(x+y)
    a^x / a^y = a^(x-y) or 1/a^(y-z)
    you should be able to wrap that up in this neat little xmas package:
    a^(11/6) b^(1/3) / c
    i don't think that the c is square rooted. the answer is posted above. but I sort know how to get there now thanks to you guys' help.

    the answer is somehow a^11/6 b^1/3 c^- 3/2, the negative -3/2 powered to base of c
    Last edited by immengin101; December 16th 2009 at 08:35 AM. Reason: more
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68
    Well, whadda heck does your denominator:
    ( squareroot(a^3 squareroot(b c)
    mean?

    You have 3 "("s but only 1 ")"
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Dec 2009
    Posts
    6
    hello, it seems that using normal typing isn't seeming to help.

    please refer to this picture for the problem.

    I was using the software algebrator to simplify this problem, however, it is giving me a different result.

    that is the solution in the brackets, however, when I insert it into algebrator, there are entirely different results.

    and I think the problem is being read by the program differently and I don't think the "a" squarerooted isn't supposed to cover the "b" that is squarerooted also. ugh, I hate this, it all sounds so confusing, I'm just trying to understand the steps.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Dec 2009
    Posts
    6
    nevermind, solved it, thanks guys!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Indice Problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 17th 2011, 04:42 AM
  2. [SOLVED] Binary Operations
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: May 10th 2010, 03:17 PM
  3. [SOLVED] Matrix Operations
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 14th 2010, 07:30 AM
  4. [SOLVED] simplification problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 24th 2009, 09:50 AM
  5. [SOLVED] bit operations...need some help
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 18th 2009, 03:05 PM

Search Tags


/mathhelpforum @mathhelpforum