# X to the xth power and other indices

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 therefore 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

Staff Emeritus
Gold Member
2)

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

1)

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
or
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.

klucar
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.

klucar
well, i am not quite sure about it so don't take it for granted...