All real number solutions of equation $\log_{2011}(2010x) = \log_{2010}(2011x)$ are in certain interval. Which one is it?

Question

This task has to be done with no calculator, but I don't have basic idea how to start. Can someone give me advice, I know this is pretty easy but I need direction for particularly this one? EDIT: I do it like this $$\log_{2011}(2010x) = \log_{2010}(2011x)$$ $${\ln{2010x}\over\ln2011}={\ln{2011x}\over\ln2010}$$ $${{\ln{2010} + \ln{x}}\over\ln2011}={\ln{2011} + \ln{x}\over\ln2010}$$ $$\ln^2{2010} + \ln{x}(\ln2010) = \ln^2{2011} + \ln{x}(\ln2011)$$ $$\ln^2{2010} - \ln^2{2011} -\ln{x}(\ln{2011} - \ln{2010}) = 0 $$ $$(\ln{2011} - \ln{2010})(\ln{2011} + \ln{2010} - \ln{x}) = 0$$ $$\ln{(2010 * 2011)\over x} = \ln{1}$$ $$x = 2010 * 2011$$ What am I doing wrong here? And also, how to even get solution (which is $x\in (0,{1\over 2011}])$

Answer 1

Hint: $$\log_{2011}(2010x) = \frac{\ln 2010 + \ln x}{\ln 2011}$$ Rewrite $\log_{2010}(2011x)$ similarly and solve for $\ln x$.

Answer 2

$ x=\frac{1}{2010 \times2011}$ Just use normal properties of log like

1) first change the base of every log to e

2) expand $ln2010x$ and $ln 2011x $ using property $ lnab=lna + lnb $

3) solve for x