Understanding why $x^{\log_b(y)} = y^{\log_b(x)}$ According to wikipedia, we have that 
$$
x^{\log_b(y)} = y^{\log_b(x)} 
$$
because
$$
x^{\log_b(y)} = b^{\log_b(x) \log_b(y)} = b^{\log_b(y) \log_b(x)} = y^{\log_b(x)}
$$
But what justifies that first leap?
$$
x^{\log_b(y)} = b^{\log_b(x) \log_b(y)}
$$

It starts from the fact that $x$ is equal to $b^{\log_b(x)}$. To elaborate, the log to the base $b$ of $x$ is defined as the number which, when $b$ is raised to this number, gives back $x$. Now, since $x = b^{\log_b(x)}$, we can always replace any occurrence of $x$ in any equation whatsoever with $b^{\log_b(x)}$. So, breaking that first leap into two steps, we first have:
$x^{log_b(y)} = (b^{\log_b(x)})^{log_b(y)}$
From here, we can apply a known rule of exponents which is that $(s^r)^t = s^{rt}$. Applying that rule we get:
$(b^{\log_b(x)})^{log_b(y)} = b^{\log_b(x)log_b(y)}$
Which is the first step in the above.
