A Bit of Algebra

In this blog post we're going to look at finding the real positive x such that: \[x^{x^3} = \frac{1}{2^{1/6}} \hspace{10 mm} (Equation 1)\]
By all means, do have a go at solving it, before reading our solution below.

The first, slightly non-obvious step is to raise both sides of the equation to the power of 3: \[ \left( x^{x^3} \right)^3 = \left( \frac{1}{2^{1/6}} \right)^3 \] Rearranging that a bit, we have: \[ x^{3x^3} = \frac{1}{2^{3/6}} \] And so: \[ \left( x^3 \right) ^{x^3} = \left( \frac{1}{2} \right) ^ {1/2} \hspace{10 mm} (Equation 2) \] If \[ a^a = b^b \] then we have a solution: \[a=b\]
So, returning to Equation 2, we have: \[x^3=1/2 \] And hence: \[x=\frac{1}{2^{1/3}} \] So, we have a solution to equation 1. But are there any others?
In fact, it is possible have \[ a \neq b \] when \[ a^a = b^b \] For example: \[ \left( \frac{1}{4} \right)^{1/4} = \left( \frac{1}{2} \right)^{1/2} \] So, returning to equation 2, we can see that another solution for x is when: \[ x^3 = \frac{1}{4} \] And hence: \[ x = \frac{1}{2^{2/3}} \] So, our solutions for x are: \[ \frac{1}{2^{1/3}} \hspace{4 mm} and \hspace{4 mm} \frac{1}{2^{2/3}} \] But are there other solutions?
If we let \[ z=x^3 \] then, returning to equation 2 we have: \[ z^z= \frac{1} {\sqrt{2}} \hspace{10 mm} \] We can plot \(z^z \) for positive z:

We can see that \(z^z\) decreases in the range 0 to 1/e, after that it increases and keeps on increasing. So for positive z, we will have a maximum of two solutions to \( z^z = k\), where k is some constant. We have two solutions already, so we now know there aren't any more,
unless of course we want to investigate negative or complex solutions...