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

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Homework Help Overview

The discussion revolves around solving the logarithmic equation \(\log_{2010} 2011x = \log_{2011} 2010x\), which involves properties of logarithms and algebraic manipulation. Participants are exploring the relationships between the logarithmic expressions and the implications for the variable \(x\).

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to manipulate the logarithmic equation by applying logarithmic identities and properties. There are discussions on transforming the equation into a form that isolates \(x\) and examining the implications of the logarithmic properties used.

Discussion Status

Several participants have provided similar approaches to the problem, with attempts to simplify the logarithmic expressions. Hints have been offered to guide the exploration of logarithmic identities, but no consensus has been reached on the final solution.

Contextual Notes

Participants are working within the constraints of the problem as presented, with no additional information provided about the context or specific requirements beyond solving the equation.

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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
 
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mafagafo said:

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).
 
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\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.
 
Glad to help. :)
 

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