Today’s problem:

  • $\frac{a}{b} = \frac{2}{3}$
  • $a^b = b^a$
  • Find $b-a$.

I’m just going to straight-up answer this below the line.

My first step would be to separate the $a$ and $b$ in the second equation, to get $a^{1/a} = b^{1/b}$. I was tempted by logs for a moment, but we don’t know that $a$ and $b$ are positive.

I can also separate the first equation to get $b = \frac{3a}{2}$.

Now I have $a^{1/a} = \left(\frac{3a}{2}\right)^{\frac{2}{3a}}$. Now we’re cooking.

Raise both sides to the power of $3a$, which is definitely the ugly thing: $a^3 = \left(\frac{3a}{2}\right)^2$ – and we’re almost there:

$4a^3 = 9a^2$, so $a = \frac{9}{4}$.

We can also say that $b-a = \frac{a}{2}$, which works out to be $\frac{9}{8}$.

(It turns out logs would have worked after all, but we didn’t know that at the time…)