For what number a > 1 is there only one solution to a^x = x^a?

e=2.7182818...

To see this, consider the function f(t)=t^(1/t). Notice that a^x=x^a if and only if a^(1/a)=x^(1/x), i.e., when f(a)=f(x). Now for positive values of t, f has a maximum value at precisely t=e. You can see this by differentiating f and setting equal to zero. Therefore f(a)=f(x) has two different solutions for x>1, except for x=e.

a=e. Consider the function f(x) = x^(1/x). Thus we have:

f(x) = e^(ln x/x)

f'(x) = e^(ln x/x) * (1 - ln x)/x²

We wish to find the maximum of f on the positive real axis. This will occur at a point where f'(x) = 0. Thus we have:

0 = e^(ln x/x) * (1 - ln x)/x²

0 = (1 - ln x)/x²

0 = 1 - ln x

ln x = 1

x = e

For x>e, we have that 1 - ln x is negative, so f'(x) is negative, so f is decreasing. For 0<x<e, 1-ln x is negative, so f'(x) is negative, so f is increasing. This means that x=e is a global maximum of f(x) on the positive real numbers, which means that for all x>0:

f(e) ≥ f(x)

e^(1/e) ≥ x^(1/x)

e^x ≥ x^e

With equality if and only if x = e. So there will be only one solution to the equation e^x = x^e. To show that there is only one such number, let a be any other number greater than 1. Since f(1) = 1 and [x→∞]lim f(x) = [x→∞]lim e^(ln x/x) = [x→∞]lim e^(1/x) = e^0 = 1, it follows from the intermediate value theorem that for any number c strictly between 1 and e^(1/e), there will exist a∈(1, e) and b∈(e, ∞) such that f(a) = f(b) = c. Further, for any a>1 such that a≠e, f(a) is strictly between 1/e. Which means that if 1<a<e, there exists b such that e<b and f(a) = f(b), and conversely if e<a, there exists b such that 1<b<e and f(a) = f(b).

In either case, if a>1 and a≠e, there exists b≠a such that f(a) = f(b), which means:

a^(1/a) = b^(1/b)

a^b = b^a

However, since a^a = a^a, a and b are thus two distinct solutions to the equation a^x = x^a. Thus, there is one and only one number a>1 for which a^x = x^a has a unique solution in the positive reals, and that is a = e. Q.E.D.

a = e will have only one solution for x iff x is restricted to the Reals, which was not explicitly stated as a requirement. For example if a = e, then one solution is

x ≈ 8.6077123504196488113 - 22.110244684283348950i

another is:

x ≈ -.63000923891390012461 + .48181368120072026243i

There will be many solutions involving the Lambert W function over the Complex numbers. Thinking beyond just algebra....

http://mathworld.wolfram.com/LambertW-Function.htm...

If we are restricted to the Reals, then probably the simplest way to solve this is to consider the function f(x) = (x^a) - (a^x) and look for it's zeros.

Four things are easily proved about this function for a > 1:

f(1) = (1-a) is negative.

Lim(x→∞) f(x) = -∞ is also negative.

f(x = a) is ALWAYS a root of f(x).

d(f(a)) / dx = (aª) * [1 - ln(a)] may be positive, negative or zero.

From these four observations we can reason as follows:

For 1 ≤ x < ∞:

If f'(a) is positive or negative there must be at least two roots. Thus the only way we can possibly have only one root is if f'(a) is zero. Then:

(aª) * (1 - ln(a)) = 0

ln(a) = 1

a = e

2

To add to what the others have said, here is another way of looking at it.

a^x = x^a implies that

a^(1/a) = x^(1/x) = y

Choosing a value of a that has only one x solution is the same as choosing a value of y that has only one x solution.

Draw the graph of x^(1/x). It has a maximum at x=e (as Pascal has shown).

So draw a horizontal line from y, and you'll see that this line intersects the curve at either

1) 2 places

2) 1 place

3) 0 places

It intersects at only one place if it intersects at the maximum point i.e. when x = a = e.

either 2 or 4:

2^4=4^2

To verify it, the easiest method would be to graph 2^x and x^2, but you probably want a more substantial explanation, so...

in order for two numbers to have and common powers, one must be a power of the other. In order for a^x to equal x^a, when the numbers are written in powers of the smaller number, they equation will be in the form

a^(a^b)=(a^b)^a

which is equivalent to:

a^(a^b)=a^ab

in order for this to be true, a^b must be equal to ab. In order for this to be true, a added to itself b times must be equal to a multiplied by itself b times. For any number greater than 2, the number multiplied by itself b times will be greater than the number added to itself b times. For 2, 2 added twice is the same as 2 multiplied twice and both are equal to 4, so 2 is the solution. However, since the equality can be reversed, 4 is also a solution, so the answers are 2 and 4.