We are only concerned with positive x. Introduce y = x³, then x^y = 3. Take the logarithm of both sides of both equations:

which simplifies to

The latter equation is trivially solved by y = 3. (We may observe in passing that the equation has no other real solutions. Indeed, the function g(y) = y·ln(y) is negative for y < 1 and, since it's derivative g'(y) = ln(y) + 1 is positive, is increasing for y > 1.)

Therefore, 3 = x³, and x = 3^1/3.

Well, we got one solution to the equation x^x3 = 3 and may consider the problem solved and the case closed. However, I was quite surprised to arrive at that solution. For, note that the value x = 3^1/3 solves, by definition, another equation, namely x³ = 3. The situations begs for a generalization. x = 3^1/3 solves

Will it also solve x^xx3 = 3? Yes, of course. Since 3 = x³ and x^x3 = 3 we can replace in the latter the exponent 3 with x³, which then asserts that indeed x^xx3 = 3.

Having made the first step, why not to continue? We can substitute x³ for the exponent 3 in the new equation, to obtain x^xxx3 = 3. Let f_n(x) = x^x...x3 with n occurrences of x, so that f₁(x) = x³ and f₂(x) = x^x3. Then, by an obvious induction, f_n(3^1/3) = 3. On the right, you can see the graphs of the first few functions f_n.

The sequence {s_n} is monotone increasing and is bounded from above. (It's a well known fact that bounded monotone sequences have a limit.) Indeed, since x > 1, s₁ > 1. We then also have

and more generally, by induction,

The sequence {s_n} is also bounded from above. To see that, first of all observe that if

then x and y are either both greater or both less than 1. If they are greater than 1 then x < y. This starts an induction: s₁ < y. Assume s_n < y. Then

We conclude that sequence {s_n} has a limit, say, A which, by necessity, satisfies

It is important to note that A may be different from y! In fact the function f(z) = z^1/z is defined for all positive z, has a maximum at z = e, is monotone increasing for z < e and is monotone decreasing for z > e.

This means that, for 1 < x < e^1/e, the equation y^1/y = x has two roots: one is less, the other is greater, than e. For example, for x = √2, (1) has two solutions: y = 2 and y = 4. (This is because 4^1/4 = 2^1/2.)

The sequence {s_n} results if one attempts to solve (1) by an iterative procedure.

where g(s) = x^s. The derivative g'(s) = dg(s)/ds is found to be