a^x = x^a

Factual Questions
1

I think there was a thread similar to this a few months ago but it was the recent thread about logarithms that made me think about this.

I’ve researched a great many websites to see why ‘e’ is so special. Oh sure, I get the usual textbook routine - it’s the basis of natural logarithms; it has a rate of growth equal to itself; it is used to compute continuously compounded interest, it is equal to (1 + (1/n))[sup]n[/sup] etc. But nothing I have ever read has actually explained what makes this number so special.

So, I tried solving the equation in the topic title (figuring ‘e’ would enter the picture sooner or later):
a[sup]x[/sup] = x[sup]a[/sup]

The equation will always have a “trivial” answer (except for zero) whenever a=x.
For example, if a=10 and x=10 then:
a[sup]10[/sup] = x[sup]10[/sup]
However, the equation will also have a second answer.

In the case of 1.5, the other answer is 7.40
For 2, the other answer is 4
For 2.5, the other answer is 2.9756
For 3 it is 2.478
For 4 it is 2
For 5 it is 1.765
For 10 it is 1.371

So, for the numbers 1.5, 2 and 2.5 the trivial answer is always lower than the other answer.
For the numbers 3, 4, 5 and 10 the trivial answer is always higher.
And where does this “shift” take place? Somewhere between 2.5 and 3. Guess that must be the number ‘e’ !!! (or 2.718281828…)
Basically, when using the number ‘e’, the equation only has one solution and that’s ‘e’. (OR for the purists among you, it has 2 solutions and they are both equal to ‘e’).
I highly doubt I am the first person to think of this but it is an interesting property of ‘e’ and one I have never seen in any calculus book. At least this shows a unique property of ‘e’ without getting into the fancy mathematical nomenclature.
It does show how the rate of “growth” changes at ‘e’. In a sense is this showing that when the number equals ‘e’, the rate of growth is equal to itself?
Okay, anyone more familiar with this? Have I done anything wrong?

2

I’ve looked at this equation before. I don’t have time to look at your secondary question right now, but here’s a way to generate solutions to your equation (the other stuff you’re interested in will hopefully fall out relatively easily from here):

(I’m copying and pasting this from another message board where I made this same post, so I’m using different variables here):

a[sup]b[/sup]=b[sup]a[/sup], and let’s assume everything’s positive so we don’t run into any trouble there. Write b=xa. Then

a[sup]xa[/sup] = x[sup]a[/sup]a[sup]a[/sup]

x[sup]a[/sup] = a[sup]a(x-1)[/sup]

Assume x is not 1 since that gives the trivial solution a=b. Take the a(x-1)th root of both sides:

a = x[sup]1/(x-1)[/sup], b = x[sup]x/(x-1)[/sup], x an arbitraty positive real number.

3

That might be the case, but to actually show that would be a lot harder than doing any of the other standard properties of e. Oh, and for a = 2, there’s a third solution.

4

Assuming b=xa, and therefore x=b/a, can we solve for a in terms of b? I can’t get any further than

a = (b/a)[sup]1/[(b/a)-1][/sup]

5

It’s a long way from “guess” to “must” … is there any actual proof that it “swaps over” at exactly e?

6

With brute force on my TI-83, I get 0.7666647, but that’s not exact.

7

I thought it was something like that. That pretty much blows a big hole in the OP’s conjecture–you have one other answer that’s bigger, but you also have one that’s smaller.

8

I actually meant -0.7666647, with a negative sign. But ultrafilter, you’re still right. Is that true for all numbers, or is it something special about the number 2?

9

Well, it’s not too surprising that you found a point where the solutions switch from a < x to a > x: given a solution a = q, x = r, with q different from r, you can find another solution a = r, x = q.

Here’s another way to look at this: You can juggle your equation around to read

(ln x) / x = (ln a) / a.

Let’s look at the function f(x) = (ln x)/x. We can see that f goes to minus infinity when x goes to zero; f goes to zero as x goes to infinity; and f(1) = 0. If we take the derivative, we find that f’(x) = (1 - ln x)/x[sup]2[/sup], which goes to zero at (hey!) x = e; it’s positive for x < e, and negative for x > e. This tells us that f(x) rises sharply from minus infinity near zero; passes through zero at x = 1; peaks at x = e, at a value of f(x) = 1/e; and then asymptotically decreases to zero as x goes to infinity. (Man, I wish I coud just post an image of the above description.)

So if we want to solve the equation f(x) = f(a) = y for some value of y, how can we do this? Well, there are four cases:
[ul]
[li] y <= 0. There’s only one possible value x and a can have in this region, so they must have the same value, somewhere in the region (0,1]. So x = a if either x or a is less than or equal to 1.[/li][li] 0 < y < 1/e. These are the interesting solutions you found. There are two values that correspond to each value of y in this region, one with 1 < x < e and the other with x > e. So we can have x = a or x different from a.[/li][li] y = 1/e. This is the only value of x for which f(x) = 1/e, so if f(x) = f(a) = 1/e, then x = a = e. This is the intersection of the two “branches” of solutions postulated in the OP.[/li][li] y > 1/e. No solutions exist.[/li][/ul]

So yes, you’re correct that there are two types of solutions. If x <= 1, the only solution for a is a = x; if x > 1, there are two solutions, the x = a solution and another one where x != a. The exception is the solution x = a = e, where these two “types” of solutions happen to coincide.

Note that what I’ve said above only works for positive values of x and a, though, since you have to assume that to get the equation f(x) = f(a) that I found above. The “extra solution” ultrafilter and TJdude825 found (involving a negative number) is only there because 2 is an integer; you can’t raise a negative number to a non-integer power without getting into complex numbers. I would expect that there would be a solution for x = {4,6,8,…} and a negative. So there will be “isolated solutions” as well, with one of the values of x or a being an even integer and the other being negative.

10

To fully solve the equation, you need Lambert’s W function, which is not for the math-phobic.

11

Some tangental info;

If you plot the curve y=x^1/x, it is undefined at 0 but increases from an infinitesimal value to a max of approx 1.444 (e^1/e) and levels off asymtotically approaching 1 as you go to infinity.

If you plot the curve x^x, it is a quasi-parabola/catenary shape with a vertex at 1/e.

12

Gawd, math confuses the crap outta me!

13

I have had a bit of a play with this particular function.
This is not going to look pretty written on a single line, but here goes.

(27/8)^(9/4) = (9/4)^(27/8)
(256/81)^(64/27) = (64/27)^(256/81)

and in general, one can find rational solutions to the relation x[sup]y[/sup]=y[sup]x[/sup] parametrically using
x=((n+1)/n)^(n+1)
y=((n+1)/n)^n

These solutions can be reflected in the line y=x. I’m not sure if that is the set of all possible positive rational solutions.

It seems to me that there ought to be a similar set of solutions in the third quadrant where both x and y are negative. It doesn’t look like there are any real solutions there because of the difficulty in raising a negative number to an irrational power.

Any good graphing package should be able to give you a picture of the relation. There are trivial solutions aloing the line y=x. And there is also a line looking roughly like one part of a hyperbola with assymptotes at x=1 and y=1 and passing through the point (e,e)

Beynod that my mathematical rigour fails. (Actually probably before that.) But it was a fun exercise to find some of the rational solutions.

14

I can’t rember how to prove it, but I seem to remeber it involves differentitaing by du/dx, where u = x^x.

15

Since a lot of folks went to the trouble of answering my question, I figured I’d use some of that information and put it on a website. (This is my free geocities website)
http://www.geocities.com/internet_web_surfer_dude/logs.htm

Maybe when I can write a tutorial on logarithms I’ll put this on my genuine website www.1728.com
As of now it’s just a very brief information summary of the message board data.

Anyway, this graph is based upon numbers generated by Cabbage’s formulas.