# Thread: [SOLVED] Logarithm with different base

1. ## [SOLVED] Logarithm with different base

Hi everyone

I need help for this problem :

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

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

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

How to deal with different base?

Thx

2. Originally Posted by songoku
Hi everyone

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

How to deal with different base?

Thx

This may help

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

3. Hi pickslides

I've tried that and still....

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

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

Next hint?

Thx

4. Are you sure this is the way the problem is stated? I've tried to solve this several ways, and the best I can come up with is

$\displaystyle \frac{x}{y}\,=\, \frac{(xy)^{log_3 \sqrt{2}}}{y} +2$

Which doesn't look too satisfying.
I can show my work, but I just wanted to post this first, and have you confirm this is really the problem. Also, has anyone else tried solving this?

QM

5. Hi QM deFuturo

Yes, that's the whole problem and that's the way it is stated. I also have tried to solve it and always get stuck...

Thx for trying

6. I conclude that this problem can't be solved

Thx