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?
$$
x^{\log_b(y)} = b^{\log_b(x) \log_b(y)}
$$
 A: 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.
A: Notice that $\def\lfrac#1#2{{\large\frac{#1}{#2}}}$$a^b=e^{b\ln(a)}$ and $\log_c (d)=\lfrac{\ln(d)}{\ln(c)}$. Therefore:
$$x^{\log_b(y)}=e^{\log_b(y)\ln(x)}=e^{\lfrac{\ln(y)}{\ln(b)}\ln(x)}=e^{\ln(y)\lfrac{\ln(x)}{\ln(b)}}=e^{\ln(y)\log_b(x)}=y^{\log_b(x)}$$
A: Their proof is kind of crappy.  Just take the log of both sides of the equation, and you'll get log(x)log(y)=log(y)log(x)
A: For an intuitive (purposefully non-rigorous) reason, consider writing logarithms in an infix notation, as $a \underline{\log} b$ rather than $\log_b a$.
Then the relation is
$$x \text{ ^ } (y \underline{\log} b) = y \text{ ^ } (x \underline{\log} b)$$
which is just one arithmetic step up from a relation you probably already know:
$$x \times (y \div b) = y \times (x \div b)$$
which itself is also just one arithmetic step up from:
$$x + (y - b) = y + (x - b)$$
It makes a nice pattern.
A: Method-1: 
Notice, we know that $\large \color{red}{\frac{1}{\log_m(n)}=\log_n(m)}$ & $\color{red}{m^{\log_m(n)}=n}$   
Now, we have
$$\large x^{\log_b(y)}=y^{\log_b(x)}$$ $$\large \iff x^{\frac{1}{\log_b(x)}}=y^{\frac{1}{\log_b(y)}}$$
$$\large \iff x^{\log_x(b)}=y^{\log_y(b)}$$
$$\large \iff b=b$$
 Method-2: we have $$\large x^{\log_b(y)}=y^{\log_b(x)}$$ taking log with base $x$ on both the sides, we get
$$\large \log_x(x^{\log_b(y)})=\log_x(y^{\log_b(x)})$$ $$\log_b(y)\log_x(x)=\log_b(x)\log_x(y)$$
$$\frac{\log_b(y)}{\log_b(x)}=\log_x(y)$$
$$\log_b(y)\cdot \log_x(b)=\log_x(y)$$
$$\log_x(y)=\log_x(y)$$
