If $2 \cdot log_e{(x -2y)} = log_e{y} + log_e{x}$, then find the numerical value of $\frac{x}{y}$

Question

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?

Answer 1

Hint :

$$2 \cdot \ln(x -2y) = \ln y + \ln x\Rightarrow \ln(x-2y)-\ln y=\ln x-\ln (x-2y)$$

i.e., $$\log \bigg(\frac{x-2y}{y}\bigg)=\log\bigg(\frac{x}{x-2y}\bigg)$$

i.e.,

$$\dfrac{x}{y}-2=\dfrac{1}{1-2\left(\dfrac{y}{x}\right)}$$ Can you complete this...?

Answer 2

Hint: Let $z = \frac xy$, dividing your last equation by $y^2$ (note that $y > 0$, as $\log y$ is defined), we get $$ \frac xy = \left(\frac xy - 2\right)^2 \iff z = (z-2)^2 $$ Now solve for $z$.

Answer 3

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$