Results 1 to 8 of 8

Math Help - Solving for x in an equation with an exponent that's a fraction?

  1. #1
    Newbie
    Joined
    May 2011
    Posts
    12

    Solving for x in an equation with an exponent that's a fraction?

    16 / (3(x-4)^(1/3))

    How would I go about solving for x in this equation? The (x-4)^(1/3) part is basically the same as the cube root of (x-4) right?


    P.S. Sorry I'm totally new at this site and haven't taken the time to learn how to use the commands so my equations are a bit of a mess
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by theant4 View Post
    16 / (3(x-4)^(1/3))

    How would I go about solving for x in this equation? The (x-4)^(1/3) part is basically the same as the cube root of (x-4) right?


    P.S. Sorry I'm totally new at this site and haven't taken the time to learn how to use the commands so my equations are a bit of a mess
    There is nothing to solve!

    There is no equal sign!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2011
    Posts
    12
    Oops! My bad... set that equal to 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,453
    Thanks
    1868
    16 / (3(x-4)^{1/3})= 0?

    That's impossible. A fraction 0 if and only if the numerator is 0. Here the numerator is 16. No value of x will make that equal to 0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    \frac{16}{3(x-4)^\frac{1}{3}}=0

    This equation has no solutions.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2011
    Posts
    12
    Okay, well here's the original equation, but it belongs in the calculus forum. I'm supposed to be finding a single critical point for it:

    f(x) = 8(x-4)^(2/3)
    and f'(x) = (2/3) 8((x-4)^(-1/3)) (1) ... hopefully I did that derivative right. Maybe that's where I'm screwing up in the first place.
    and that equation can be set up as 16 / (3(x-4)^(1/3)... I'm not sure which would be easier to solve.

    And then in all the other equations we've done you would solve for x by setting the left side to 0... so basically I have no clue what to do.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    You are correct that the euqation has no solution, but don't forget places where the derivative is undefined are also critical points.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    No, you are not supposed to find an expression and set it equal to zero. You are supposed to find critical points of the function and gain some understanding.

    The Function - You should notice that if this were an integer exponent it would cause you little problem. The complication is the fractional exponent. Okay, why is that different? How does it act differently? Is there a Domain difference? If it were 3/2, would that be different from 2/3? Have you tried to draw it? There is a subtle problem if you want some CAS to draw it for you. Do you know this? You should find that x^{\frac{2}{3}} is NOT the same as \sqrt[3]{x^{2}}. Why would that be?

    The First Derivative - Why don't we see what we can tell about the function itself before we get into this?
    Last edited by TKHunny; May 9th 2011 at 08:44 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Negative fraction with exponent?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 28th 2011, 08:54 PM
  2. fraction exponent question
    Posted in the Algebra Forum
    Replies: 5
    Last Post: December 9th 2010, 05:42 PM
  3. Solving f(x) equation with a fraction
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 20th 2010, 04:00 PM
  4. Help with turning a fraction into a exponent
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 3rd 2009, 09:51 PM
  5. fraction exponent in denominator
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 14th 2008, 02:13 PM

Search Tags


/mathhelpforum @mathhelpforum