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Logarithmic equation

  1. Apr 12, 2014 #1
    1. The problem statement, all variables and given/known data
    [tex]\log_{2010} 2011x = \log_{2011} 2010x[/tex]

    3. The attempt at a solution
    [tex]\log_{2010} 2011x = \log_{2011} 2010x[/tex]
    [tex]\frac{\log 2011x}{\log 2010} = \frac{\log 2010x}{\log 2011} [/tex]
    [tex]\frac{\log 2011x}{\log 2010} - \frac{\log 2010x}{\log 2011} = 0[/tex]
    [tex]\frac{\log 2011\log 2011x - \log 2010 \log 2010x}{\log 2010 \log 2011} = 0[/tex]
    [tex]\log 2011\log 2011x - \log 2010 \log 2010x = 0[/tex]

    What's next?

    Answer should be [tex] x = 1/4042110 [/tex]
     
  2. jcsd
  3. Apr 12, 2014 #2
    Hi mafagafo!

    Hint: Use log(a*b)=log(a)+log(b).
     
  4. Apr 12, 2014 #3
    [tex]\log 2011\log 2011x - \log 2010 \log 2010x = 0[/tex]
    [tex]\log 2011 (\log 2011 + \log x) - \log 2010 (\log 2010 + \log x) = 0[/tex]
    [tex](\log 2011 )^2 + \log 2011 \log x - ((\log 2010 )^2 + \log 2010 \log x) = 0[/tex]
    [tex](\log 2011 )^2 + \log 2011 \log x - (\log 2010 )^2 - \log 2010 \log x = 0[/tex]
    [tex]\log 2011 \log x - \log 2010 \log x = (\log 2010 )^2 - (\log 2011 )^2[/tex]
    [tex]\log x \cdot (\log 2011 - \log 2010) = (\log 2010 )^2 - (\log 2011 )^2[/tex]
    [tex]\log x = \frac{(\log 2010 - \log 2011)(\log 2010 + \log 2011)}{\log 2011 - \log 2010}[/tex]
    [tex]\log x = \frac{(\log 2010 - \log 2011)(\log 2010 + \log 2011)}{-(\log 2010 - \log 2011)}[/tex]
    [tex]\log x = - (\log 2010 + \log 2011) = - \log 4042110 = \log (4042110^{-1}) [/tex]
    [tex]x = 4042110^{-1} = \frac{1}{4042110} [/tex]

    Thank you, Pranav-Arora.
     
  5. Apr 12, 2014 #4
    Glad to help. :)
     
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