How to solve this system of logarithm equations?

Question

$x^{log_8(y)}+y^{log_8(x)}=4\\log_4(x)-log_4(y)=1$

I have no idea how to start this.

Answer 1

$$x^{log_8(y)}+y^{log_8(x)}=4$$ $$y^{log_8(x)}+y^{log_8(x)}=4$$ $$2y^{log_8(x)}=4$$ $$y^{log_8(x)}=2$$ $$log_8(x)=log_y (2)$$ 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:

1. $a^{log_b(c)}=c^{log_b(a)}$
2. $log_a b=\frac{1}{log_b a}$

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.

Answer 2

HINT:

$$\begin{cases} x^{\log_8(y)}+y^{\log_8(x)}=4\\ \log_4(x)-\log_4(y)=1 \end{cases}\Longleftrightarrow$$

$$\begin{cases} x^{\log_8(y)}+y^{\log_8(x)}=4\\ \frac{\ln(x)}{\ln(4)}-\frac{\ln(y)}{\ln(4)}=1 \end{cases}\Longleftrightarrow$$

$$\begin{cases} x^{\log_8(y)}+y^{\log_8(x)}=4\\ \frac{\ln(x)-\ln(y)}{\ln(4)}=1 \end{cases}\Longleftrightarrow$$

$$\begin{cases} x^{\log_8(y)}+y^{\log_8(x)}=4\\ \ln(x)-\ln(y)=\ln(4) \end{cases}\Longleftrightarrow$$

$$\begin{cases} x^{\log_8(y)}+y^{\log_8(x)}=4\\ \frac{x}{y}=4 \end{cases}\Longleftrightarrow$$

$$\begin{cases} x^{\frac{\ln(y)}{\ln(8)}}+y^{\frac{\ln(x)}{\ln(8)}}=4\\ \frac{x}{y}=4 \end{cases}\Longleftrightarrow$$

$$\begin{cases} 2x^{\frac{\ln(y)}{\ln(8)}}=4\\ \frac{x}{y}=4 \end{cases}\Longleftrightarrow$$

$$\begin{cases} x^{\frac{\ln(y)}{\ln(8)}}=2\\ \frac{x}{y}=4 \end{cases}\Longleftrightarrow$$

$$\begin{cases} \frac{\ln(2)}{\ln(x)}=\frac{\ln(y)}{\ln(8)}\\ \frac{x}{y}=4 \end{cases}\Longleftrightarrow$$

$$\begin{cases} \ln(2)\ln(8)=\ln(y)\ln(x)\\ \frac{x}{y}=4 \end{cases}\Longleftrightarrow$$

$$\begin{cases} \ln(2)\ln(8)=\ln(y)\ln(x)\\ x=4y \end{cases}\Longleftrightarrow$$

$$\begin{cases} \ln(2)\ln(8)=\ln(y)\ln(4y)\\ x=4y \end{cases}\Longleftrightarrow$$

$$\begin{cases} \ln(4)\ln(y)+\ln^2(y)=\ln(2)\ln(8)\\ x=4y \end{cases}\Longleftrightarrow$$

$$\begin{cases} \ln(4)\ln(y)+\ln^2(y)+\frac{\ln^2(4)}{4}=\ln(2)\ln(8)+\frac{\ln^2(4)}{4}\\ x=4y \end{cases}\Longleftrightarrow$$

$$\begin{cases} \left(\ln(y)+\frac{\ln(4)}{2}\right)^2=\ln(2)\ln(8)+\frac{\ln^2(4)}{4}\\ x=4y \end{cases}\Longleftrightarrow$$

$$\begin{cases} \ln(y)+\frac{\ln(4)}{2}=\pm\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}\\ x=4y \end{cases}\Longleftrightarrow$$

$$\begin{cases} \ln(y)=\pm\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}-\frac{\ln(4)}{2}\\ x=4y \end{cases}\Longleftrightarrow$$

$$\begin{cases} y=e^{\pm\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}-\frac{\ln(4)}{2}}\\ x=4y \end{cases}\Longleftrightarrow$$

$$\begin{cases} y_1=e^{\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}-\frac{\ln(4)}{2}} \vee y_2=e^{-\sqrt{\ln(2)\ln(8)+\frac{\ln^2(4)}{4}}-\frac{\ln(4)}{2}}\\ x=4y \end{cases}\Longleftrightarrow$$ $$\begin{cases} y_1=\frac{1}{8} \vee y_2=2\\ x=4y \end{cases}\Longleftrightarrow$$

$$\begin{cases} y_1=\frac{1}{8} \vee y_2=2\\ x_1=\frac{1}{2} \vee x_2=8 \end{cases}$$

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We got 4 solutions:

$$x_1=\frac{1}{2},y_1=\frac{1}{8}$$ $$x_2=8,y_2=2$$