<s>[INST] If $2 \cdot log_e{(x -2y)} = log_e{y} + log_e{x}$, then find the numerical value of $\frac{x}{y}$ If $2 \cdot log_e{(x -2y)} = log_e{y} + log_e{x}$, then find the numerical value of $\frac{x}{y}$
My try:

$2 \cdot log_e{(x -2y)} = log_e{y} + log_e{x}$
$log_e{(x-2y)^2} = log_e{xy}$
$(x - 2y)^2 = xy$

But expanding the LHS doesn't give the result. What should I do further?
 [/INST] If you are asked to calculate $\left(\frac {x}{y} \right) $ start with that and assume it to $t$.
Then, $2 \cdot log_e{(x -2y)} = log_e{y} + log_e{x}$ becomes
$2 \cdot log_e{(ty -2y)} = log_e{y} + log_e{ty}$
$\implies$${y^2}{{(t-2)}^2}=ty^2$
$\implies$${t^2}-4t+4=t$
$\implies$$t^2-5t+4=0$
$\implies$$(t-4)(t-1)=0$
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