A Bit of Algebra

In this blog post we're going to look at finding the real positive x such that: $$x^{x^3}=\frac{1}{2^{1/6}}(Equation1)$$
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: $${\left(x^3\right)}^{x^3}={\left(\frac{1}{2}\right)}^{1/2}(Equation2)$$ 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 $$a\ne b$$ 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: $$\frac{1}{2^{1/3}}and\frac{1}{2^{2/3}}$$ But are there other solutions?
If we let $$z=x^3$$ then, returning to equation 2 we have: $$z^z=\frac{1}{\sqrt{2}}$$ 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...