# Possible to solve x^a=a^x for a?

1. Jun 26, 2010

### maxbashi

actually for x, but it probably doesnt matter either way.

I was hoping to find an answer for x in terms of A just out of curiosity, to see where the points on these graphs intersect other than at x=a (x^2 and 2^x meet at both x=2 and x=4 for example). Is it solvable?

Last edited: Jun 27, 2010
2. Jun 27, 2010

### JJacquelin

Hello !
Not in terms of an elementary function, Yes in terms of a special function : the Lambert's W(X) function.
a = -x W(X)/ln(x) where X = -ln(x)/x
The Lambert's W function is a multiform function. So, more than one root may be obtained in some cases.

3. Jun 27, 2010

### Gregg

$$e^{x \ln a} = x^a$$

$$-1 = -x e^{-\frac{x \ln a}{a}}$$

$$-\frac{\ln a}{a} = -\frac{x \ln a}{a} e^{-\frac{x \ln a}{a}}$$

I think this is the way to do it!

4. Jun 27, 2010

### HallsofIvy

No, it's not. You have just moved things around without solving for a.

5. Jun 27, 2010

### jackmell

I think Greg was just getting it into Lambert W form after which the answer is immediate except I think we should have a $1^{1/x}$ term in there when we take the $1/x$ root (or 1/a root if solving for x). So that I get:

$$a=-\frac{x}{\log(x)}\textbf{W}\left[-e^{2k\pi i/x} \frac{\log(x)}{x}\right],\quad k=0,\pm 1, \pm 2, \cdots$$

6. Jun 28, 2010

### Gregg

Hi could you talk through $$-e^{2k\pi i/x}$$ ? I think I know why you have done that and my rearranging gives one of your solutions

7. Jun 28, 2010

### arildno

You cannot, as such, solve for "a" in terms of "x" or vice versa.

However, it is possible to get an explicit expression for the curve(s) traced out in the first quadrant of the x-a-plane, as given by your equation.

Set:
$$x=t^{k}, a=t^{m}$$
Thus, inserting this in your equation, we get:
$$t^{kt^{m}}=t^{mt^{k}}\to{kt^{m}}=mt^{k}$$
This can be rearranged as:
$$t^{k-m}=\frac{k}{m}$$
Here, one branch of the solution curve is given by k=m.
Fopr the others, we get:
$$t=(\frac{k}{m})^{\frac{1}{k-m}}$$
whereby we get:
$$x=(\frac{k}{m})^{\frac{k}{k-m}}\to{x}=c^{\frac{c}{c-1}},c=\frac{k}{m}(1)$$
$$a=c^{\frac{1}{c-1}}(2)$$

Thus, for $c\neq{1}[/tex], we get the parametric representation of solutions (x,a) by (1) and (2). 8. Jun 28, 2010 ### jackmell I just know enough to get in trouble: $$x^a=a^x$$ $$e^{a\log(x)}=a^x$$ $$1=a^xe^{-a\log(x)}$$ at this point we take the [itex]x'th[/tex] root: $$1^{1/x}=ae^{-a\log(x)/x}$$ but it seems to me that [itex]1^{1/x}=e^{i/x(2k\pi)},\quad k=0,\pm 1,\pm 2, \cdots$

then just multiplying both sides by $-\log(x)/x$ gets it into Lambert W form and the expression for a then becomes immediate.

Really though this is a little unclear to me: The W function is infinitely valued, the log is infinitely valued, and then the $1^{1/x}$ is multi-valued. It would take me some time to really wade through all this and get a good handle on it.

9. Jun 28, 2010

### jackmell

What's wrong with:

$$a=-\frac{x}{\log(x)}\textbf{W}\left[-e^{2k\pi i/x} \frac{\log(x)}{x}\right],\quad k=0,\pm 1, \pm 2, \cdots$$

10. Jun 28, 2010

### arildno

I should have qualified my statement by "solvable in terms of elementary functions",

11. Jun 29, 2010

### JJacquelin

Arildno wrote :
x = function of parametrer (c)
a = another function of (c).
This isn't the solution of equation x^a=a^x. In fact, this is a parametric relationship between x and a. The usual equation is replaced by a parametric equation without solving it.
In order to get to the solution of the equation, still it would be necessary to express c as a function of x, which can be done thanks to the Lambert W function. Finally the same result would be obtained that the result already known, which anyways includes the Lambert W function.

12. Jun 29, 2010

### arildno

The pairs of numbers (x,a) that solve the equation are given a parametric representation in terms of "c".
and later on qualified that to mean unsolvable in terms of elementary functions.

The original equation can be regarded as an implicit representation of a curve in the x-a-plane, I have given an explicit representation of that solution curve.

Last edited: Jun 29, 2010
13. Jun 29, 2010

### JJacquelin

OK, of course I agree.
My message was not a reproach, but just to make even more clair that turning the equation into parametric form isn't solving it. Sorry if my way to express it may throw into confusion.
Cordially

14. Jun 29, 2010

### arildno

That's okay.

And, I agree:
Giving an explicit expression for the solution curve, as I did, is not generally what people think of as "solving" an equation.

The parametric representation does, however, make it trivial to graph the curve, and analytically, it is easy to see that the solution curve is a trident, with its joint at (e,e).

Consistent with this, the representation proves that for 0<=x<=1, only unique solutions exist, along the line segment x=a

Last edited: Jun 29, 2010
15. Jun 29, 2010

### Dickfore

$$x^{a} = a^{x}, \ x, a > 0$$

Take the logarithm of both sides:

$$a \ln x = x \ln a$$

$$\frac{\ln x}{x} = \frac{\ln a}{a}$$

Define the function:

$$f(x) \equiv \frac{\ln x}{x}, \ x > 0$$

The inverse of f is a multi-valued function which can be expressed in terms of the Lambert W-function:

$$W(z) e^{W(z)} = z$$

Namely, taking $$w = -\ln t$$, we get:

$$w e^{w} = -\ln t \, e^{-\ln t} = -\frac{\ln t}{t}$$

So, we can finally write:

$$f(t) = -w e^{w}, w = -\ln t, \ t = e^{-w}$$

and the inverse is represented as $f^{-1}(x) \equiv e^{-W(-x)}$. As we said, there are many branches of the W-function. In general, if we have an equation of the form:

$$f(x) = f(y)$$

and $f^{-1}$ has several branches, then we can write:

$$x = f^{-1}_{k}[f(y)]$$

One of these solutions is always $x = y$. So, in your case, we have:

$$x = \exp\left[-W \left(-\frac{\ln a}{a}\right)\right]$$