How to solve this system of logarithm equations? $x^{log_8(y)}+y^{log_8(x)}=4\\log_4(x)-log_4(y)=1$
I have no idea how to start this.
 A: $$x^{log_8(y)}+y^{log_8(x)}=4$$
$$y^{log_8(x)}+y^{log_8(x)}=4$$
$$2y^{log_8(x)}=4$$
$$y^{log_8(x)}=2$$
$$log_8(x)=log_y (2)$$
Again, we have $$log_4(x)-log_4(y)=1$$
$$log_4(\frac{x}{y})=1$$
$$(\frac{x}{y})=4^1$$
$$x=4y$$
Hence from first relation we have,
$$log_{2^3}(4y)=log_y (2)$$
$$\frac{1}{3}(2+log_2 (y))=log_y (2)$$
Say $log_y (2)=a$ 
Above equation becomes
$$\frac{1}{3}(2+\frac{1}{a})=a$$
$$3a^2-2a-1=o$$
$$3a^2-3a+a-1=o$$
$$(3a+1)(a-1)=0$$
Hence $a=1$ or $-\frac{1}{3}$ 
For $a=1$ 
So, $y=2$ and $x=8$ 
And for $a=-\frac{1}{3}$ 
$y=\frac{1}{8}$ and $x=\frac{1}{2}$
FORMULAE USED: 


*

*$a^{log_b(c)}=c^{log_b(a)}$

*$log_a b=\frac{1}{log_b a}$


EDIT: 
1. Proof:  Say $a^{log_b(c)}= p$ 
$$log_b (a^{log_b(c)})= log_b p$$
$$log_b c \cdot log_b a = log_b p$$
$$log_b a \cdot log_b c = log_b p$$
$$log_b (c^{log_b(a)}) = log_b p$$
$$c^{log_b(a)}=p$$
Hence proved.
A: HINT:
$$\begin{cases}
x^{\log_8(y)}+y^{\log_8(x)}=4\\
\log_4(x)-\log_4(y)=1
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
x^{\log_8(y)}+y^{\log_8(x)}=4\\
\frac{\ln(x)}{\ln(4)}-\frac{\ln(y)}{\ln(4)}=1
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
x^{\log_8(y)}+y^{\log_8(x)}=4\\
\frac{\ln(x)-\ln(y)}{\ln(4)}=1
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
x^{\log_8(y)}+y^{\log_8(x)}=4\\
\ln(x)-\ln(y)=\ln(4)
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
x^{\log_8(y)}+y^{\log_8(x)}=4\\
\frac{x}{y}=4
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
x^{\frac{\ln(y)}{\ln(8)}}+y^{\frac{\ln(x)}{\ln(8)}}=4\\
\frac{x}{y}=4
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
2x^{\frac{\ln(y)}{\ln(8)}}=4\\
\frac{x}{y}=4
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
x^{\frac{\ln(y)}{\ln(8)}}=2\\
\frac{x}{y}=4
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
\frac{\ln(2)}{\ln(x)}=\frac{\ln(y)}{\ln(8)}\\
\frac{x}{y}=4
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
\ln(2)\ln(8)=\ln(y)\ln(x)\\
\frac{x}{y}=4
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
\ln(2)\ln(8)=\ln(y)\ln(x)\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
\ln(2)\ln(8)=\ln(y)\ln(4y)\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
\ln(4)\ln(y)+\ln^2(y)=\ln(2)\ln(8)\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
\ln(4)\ln(y)+\ln^2(y)+\frac{\ln^2(4)}{4}=\ln(2)\ln(8)+\frac{\ln^2(4)}{4}\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
\left(\ln(y)+\frac{\ln(4)}{2}\right)^2=\ln(2)\ln(8)+\frac{\ln^2(4)}{4}\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
\ln(y)+\frac{\ln(4)}{2}=\pm\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
\ln(y)=\pm\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}-\frac{\ln(4)}{2}\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
y=e^{\pm\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}-\frac{\ln(4)}{2}}\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
y_1=e^{\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}-\frac{\ln(4)}{2}} \vee y_2=e^{-\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}-\frac{\ln(4)}{2}}\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
y_1=\frac{1}{8} \vee y_2=2\\
x=4y
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
y_1=\frac{1}{8} \vee y_2=2\\
x_1=\frac{1}{2} \vee x_2=8
\end{cases}$$

We got 4 solutions:
$$x_1=\frac{1}{2},y_1=\frac{1}{8}$$
$$x_2=8,y_2=2$$
