Logaritmic equation $\log_2 \log_x(x-3y)=-1$

Question

Solve system of equations

$\log_2 \log_x(x-3y)=-1$

$x\times y^{\log_x y}=y^{\frac{5}{2}}$

I managed to reduce second equation to

$\frac{1}{\log_x y} +\log_x y =\frac{5}{2}$

But I dont know what to do with first one.

Answer 1

$\log_2 log_x(x-3y)=-1$

$\log_x(x-3y)=2^{-1}$

$x-3y=x^\frac{1}{2}$

$x-\sqrt x=3y$

Using your attempts

$\frac{1}{log_x y} +log_x y =\frac{5}{2}$

Let $t=\log_x y$,then,$2t^2-5t+2=0$,$t=2$ and $y=x^2$,or, $t=\frac{1}{2}$ and $x=y^2$

$x\in R^+-\left\{1\right\}$,$y\in R^+$,$x-3y>0$

Answer 2

$$\frac{1}{\log_x y} +\log_x y =\frac{5}{2}$$ set $t=\log_x y$ $$\frac{1}{t}+t=\frac{5}{2}$$ $t_1=1/2;t_2=2$ $$\log_x y=\frac{1}{2}\quad\quad(1)$$

Back to the first equation $$\log _2\left(\log _x(x-3 y)\right)=-1$$ means $$\log _x(x-3 y)=\frac{1}{2}$$ Comparing with $(1)$ we see that $$x-3y=y\to x = 4y$$ Plug in $(1)$ $$\log_{4y}\,y=\frac{1}{2}$$ $$y=(4y)^{1/2}\to y^2=4y\to y =4$$ Therefore $x=16,y=4$ is the solution of the system.