Solve Logarithmic Equation: \log_{2010} 2011x = \log_{2011} 2010x

[mafagafo]

Homework Statement

\log_{2010} 2011x = \log_{2011} 2010x

The Attempt at a Solution

\log_{2010} 2011x = \log_{2011} 2010x
\frac{\log 2011x}{\log 2010} = \frac{\log 2010x}{\log 2011}
\frac{\log 2011x}{\log 2010} - \frac{\log 2010x}{\log 2011} = 0
\frac{\log 2011\log 2011x - \log 2010 \log 2010x}{\log 2010 \log 2011} = 0
\log 2011\log 2011x - \log 2010 \log 2010x = 0

What's next?

Answer should be x = 1/4042110

 

[Saitama]

Homework Statement

\log_{2010} 2011x = \log_{2011} 2010x

The Attempt at a Solution

\log_{2010} 2011x = \log_{2011} 2010x
\frac{\log 2011x}{\log 2010} = \frac{\log 2010x}{\log 2011}
\frac{\log 2011x}{\log 2010} - \frac{\log 2010x}{\log 2011} = 0
\frac{\log 2011\log 2011x - \log 2010 \log 2010x}{\log 2010 \log 2011} = 0
\log 2011\log 2011x - \log 2010 \log 2010x = 0

What's next?

Answer should be x = 1/4042110

Hi mafagafo!

Hint: Use log(a*b)=log(a)+log(b).

 

[mafagafo]

\log 2011\log 2011x - \log 2010 \log 2010x = 0
\log 2011 (\log 2011 + \log x) - \log 2010 (\log 2010 + \log x) = 0
(\log 2011 )^2 + \log 2011 \log x - ((\log 2010 )^2 + \log 2010 \log x) = 0
(\log 2011 )^2 + \log 2011 \log x - (\log 2010 )^2 - \log 2010 \log x = 0
\log 2011 \log x - \log 2010 \log x = (\log 2010 )^2 - (\log 2011 )^2
\log x \cdot (\log 2011 - \log 2010) = (\log 2010 )^2 - (\log 2011 )^2
\log x = \frac{(\log 2010 - \log 2011)(\log 2010 + \log 2011)}{\log 2011 - \log 2010}
\log x = \frac{(\log 2010 - \log 2011)(\log 2010 + \log 2011)}{-(\log 2010 - \log 2011)}
\log x = - (\log 2010 + \log 2011) = - \log 4042110 = \log (4042110^{-1})
x = 4042110^{-1} = \frac{1}{4042110}

Thank you, Pranav-Arora.

 

[Saitama]

Glad to help. :)