1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

X to the xth power and other indices

  1. Aug 24, 2003 #1


    User Avatar
    Science Advisor
    Gold Member

    Recently I've considering the two functions xy= yx and y = xx.

    1) For xy= yx can y be found in terms of x? I'm sure I've seen such a solution before.

    2) In y = xx when y (and therfore x)is postive what value of y gives the minium value for x? I know it's rougly 0.45 and that dy/dx = 0 at this point, but it's a long time since I've done more advanced differentiation and I don't even know if you can differentiate xx
  2. jcsd
  3. Aug 24, 2003 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


    y(x) = xx
    ln y(x) = x ln x

    (1 / y(x)) * y'(x) = 1 * ln x + x (1 / x)
    y'(x) = y(x) (1 + ln x)
    y'(x) = xx(1 + ln x)

    Since xx > 0, y'(x) can be 0 iff
    1 + ln x = 0
    ln x = -1
    x = 1/e

    So the minimum of xx occurs at x = 1/e = 0.368


    y = x

    Ok, ok, that's not the only solution. I highly doubt there is an elementary solution for this, but I can get you started:

    for x, y > 0:

    xy = yx
    y ln x = x ln y
    y / ln y = x / ln x
    ln y / y = ln x / x

    (note each step is reversible)

    so consider f(t) = ln t / t
    then f'(t) = (1 - ln t) / t2

    So we see that f(t) is strictly increasing for t < e and strictly decreasing for t > e. This means that for any z, f(t)=z has at most two solutions. More specifically one can fairly easily show that:

    the equation f(t) = z has:
    exactly 1 solution if z <= 0
    exactly 2 solutions if 0 < z < 1/e
    exactly 1 solution if z = 1/e
    exactly 0 solutions if z > 1/e

    We've seen that if xy = yx iff f(x)=f(y).

    y = x is clearly a solution... meaning that it is the only solution (for a given x) iff f(x) <= 0 or f(x) = 1/e... that is if x <= 1 or x = e.

    Also, we can see that if there are two solutions of xy = yx for a given x, then either
    1 < x < e < y
    1 < y < e < x

    For example, the only nontrivial solution I know off the top of my head is 24 = 42... clearly 1 < 2 < e < 4

    But as I mentioned, I strongly suspect no elementary closed form solution is possible.
  4. Aug 26, 2003 #3
    I believe it is

    y(x) = (x productlog(-ln(x)/x))/ln(x);
    where prductlog(x) gives the principal solution for w in x=w e^w.
    Not all solutions are found.
  5. Aug 27, 2003 #4
    well, i am not quite sure about it so dont take it for granted...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook