Results 1 to 3 of 3

Math Help - Exp of Log simplification

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    6

    Exp of Log simplification

    Nice to find this forum!

    Could someone help here?

    I've this function:

    <br />
f(u,v) = e^{[\ln(v)^{-1/\beta}\ln(u)^{-1/\beta}]^{-\beta}}<br />

    that I need to simplify in a polynomial form with neither
    exp nor log but with all the exponents in evidence.

    Really like:

    <br />
f(u,v) = [polynom(v)^{-1/\beta}\ ...\ polynom(u)^{-1/\beta}]^{-\beta}<br />

    Is that possible? Any suggestion?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by mamboking View Post
    Nice to find this forum!

    Could someone help here?

    I've this function:

    <br />
f(u,v) = e^{[\ln(v)^{-1/\beta}\ln(u)^{-1/\beta}]^{-\beta}}<br />

    that I need to simplify in a polynomial form with neither
    exp nor log but with all the exponents in evidence.

    Really like:

    <br />
f(u,v) = [polynom(v)^{-1/\beta}\ ...\ polynom(u)^{-1/\beta}]^{-\beta}<br />

    Is that possible? Any suggestion?

    Thanks!
    I dont think this can be done, sorry to say, you could do fairly well except what to do with

    e^{ln(x)^{\frac{-1}{\beta}}}

    which to my knowledge cannot be simplified...I might be missing something simple though
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by mamboking View Post
    Nice to find this forum!

    Could someone help here?

    I've this function:

    <br />
f(u,v) = e^{[\ln(v)^{-1/\beta}\ln(u)^{-1/\beta}]^{-\beta}}<br />

    that I need to simplify in a polynomial form with neither
    exp nor log but with all the exponents in evidence.

    Really like:

    <br />
f(u,v) = [polynom(v)^{-1/\beta}\ ...\ polynom(u)^{-1/\beta}]^{-\beta}<br />

    Is that possible? Any suggestion?

    Thanks!
    I'm not sure how to interpret this part:

    \ln(v)^{-\frac{1}{\beta}}

    is it: \ln(v^{-\frac{1}{\beta}}) or (\ln(v))^{-\frac{1}{\beta}} ??
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. I don't get this simplification
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 28th 2011, 05:59 PM
  2. Simplification
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 1st 2010, 05:52 AM
  3. Simplification
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 16th 2009, 02:20 PM
  4. Simplification
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 29th 2008, 05:14 AM
  5. Simplification
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 3rd 2008, 11:04 PM

Search Tags


/mathhelpforum @mathhelpforum