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. :)