Logarithm with Different Base: Solving for x/y

[songoku]

Homework Statement

Hi everyone

I need help for this problem :

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

Homework Equations

\log_bx = \frac{\log_ax}{\log_ab}

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 :)

 

[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)

 

[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