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.
Method-1: 
Notice, we know that $\large \color{red}{\frac{1}{\log_m(n)}=\log_n(m)}$ &amp; $\color{red}{m^{\log_m(n)}=n}$   

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
$$\large \log_x(x^{\log_b(y)})=\log_x(y^{\log_b(x)})$$ $$\log_b(y)\log_x(x)=\log_b(x)\log_x(y)$$
$$\frac{\log_b(y)}{\log_b(x)}=\log_x(y)$$
$$\log_b(y)\cdot \log_x(b)=\log_x(y)$$
$$\log_x(y)=\log_x(y)$$