$a^b=b$; $a \ne b$ Find a rational solution. I am a high school teacher and last week we were given some example problems for the new End of Course Exam for Algebra 1.
One such is "$a^b=b$; $a \ne b$ Find a rational solution."
I know I can take $a=b^{\frac{1}{b}}$ and use any value of $b=\frac{1}{k}$ where $k$ is a positive integer greater than 1. to find $a$. So if I let $k=5$ then $(a,b)=(\frac{1}{5^5},\frac{1}{5})=(\frac{1}{3125},\frac{1}{5})$
Is there a more intuitive way to solve this that I am not seeing?
 A: OK, I started by expressing $b = \frac{x}{y}$, where $x$ and $y$ integers with no common factors, $x > 0$ and $y > 0$.  (We can assume $b$ is not negative unless we're going to deal with imaginary numbers etc.).
This means that $a = (\frac{x}{y})^{\frac{y}{x}}$, or written differently: $a = (\sqrt[x]{\frac{x}{y}})^y$
Now, since $x$ and $y$ share no common factors, for $a$ to be rational, we need $\sqrt[x]{\frac{x}{y}} = \frac{\sqrt[x]{x}}{\sqrt[x]{y}}$ to be rational.  Again, since they share no common factors, this means we need both $\sqrt[x]{x}$ and $\sqrt[x]{y}$ to be rational.
So: when is $\sqrt[x]{x}$ rational (for integer $x$)? It doesn't work for $x > 1$ (the smallest square number is > 2, the smallest cube number is > 3, etc, so the $x^{th}$ root of $x$ is never going to be rational.  This means that your only possible value for $x$ is 1.
Given $x = 1$, any $y$ you choose is rational.  So, you wind up at exactly your solution (where $y = k$):
$$b = \frac{1}{y},  a = (\frac{1}{y})^y$$
I don't know whether this is more or less intuitive, but I believe this shows that your family of solutions is the only one.

As mentioned in the comments, it might not be obvious that $\frac{\sqrt[x]{x}}{\sqrt[x]{y}}$ being rational implies $\sqrt[x]x$ and $\sqrt[x]y$ are rational.  It only works because we're looking at roots of co-prime integers:
Suppose $\frac{\sqrt[z]{x}}{\sqrt[z]{y}}$ is rational, but $\sqrt[z]y$ is not (for integer $z$, and co-prime integers $x$ and $y$).
Dissolve $y$ into prime factors: $y = {p_1^{m_1}}{p_2^{m_2}}...$  - for $\sqrt[z]y$ to be irrational, we must have at least one $m_k$ that is not a multiple of $z$.
However, since $\frac{\sqrt[z]{x}}{\sqrt[z]{y}}$ is rational, when we factorise $x$ into primes, then we must end up with $x = q_1^{n_1}q_2^{n_1}...{p_k^{m_k}}$, so that the $p_k^{m_k}$ is "cancelled out".  However, this implies a common factor between $x$ and $y$, which is a contradiction.
