Calculate the ratio x/y if 2log5(x-3y)=log5(2x)+log5(2y)
$\displaystyle 2\log_{5}(x-3y)=\log_{5}(2x)+\log_{5}(2y)$
$\displaystyle \log_{5}(x-3y)^2=\log_{5}(4xy)$
$\displaystyle (x-3y)^2=4xy$
Now, devide both sides by 4x to obtain a formula for y.
and then deivde both sides by 4y to obtain a formula for x.
Hence, You got a formula for x and a formula for y.
Then you are able to find the ratio $\displaystyle \frac{x}{y}$.
First note that $\displaystyle x,y > 0$. Now,
$\displaystyle 2 \log_5 (x - 3y) = \log_5 (2x) + \log_5 (2y)$
$\displaystyle \Rightarrow \log_5 (x - 3y)^2 = \log_5 (4xy)$
$\displaystyle \Rightarrow (x - 3y)^2 = 4xy$
$\displaystyle \Rightarrow x^2 - 10xy + 9y^2 = 0$
$\displaystyle \Rightarrow (x - 9y)(x - y) = 0$
So that $\displaystyle x - 9y = 0 \implies \boxed{ \frac xy = 9}$
or
$\displaystyle x - y = 0 \implies \boxed{\frac xy = 1}$