Logarithm with different base

1. Aug 13, 2009

songoku

1. The problem statement, all variables and given/known data
Hi everyone

I need help for this problem :

If $$2*\log_2 (x-2y)=\log_3 (xy)$$ , find $$\frac{x}{y}$$

2. Relevant equations
$$\log_bx = \frac{\log_ax}{\log_ab}$$

3. The attempt at a solution
$$2*\log_2 (x-2y)=\log_3 (xy)$$

$$\log_2 (x-2y)^2=\log_3 (xy)$$

$$\frac{\log_2 (x-2y)^2}{\log_2 2}=\frac{\log_2 (xy)}{\log_2 3}$$

$$\log_2 (x-2y)^2 * \log_2 3 = \log_2 (xy)$$

Then, I stuck .....

Thx :)

2. Aug 14, 2009

rock.freak667

Are you sure you wrote the question down correctly? I ask because log23 is not an integer and you be basically computing (x-2y)k=xy where k is not an integer (so you'd not be able to find x/y)

3. Aug 14, 2009

songoku

Hi rock.freak667

At least that's the whole question that was given to me by my friend. Your post assure me that this question can't be solved.

Thx a lot for pointing out that $$\log_2 3$$ is not an integer