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Algebra Level 3

( log 6 3 ) 2 + 1 log 2 6 log 18 6 = ? \left ( \log_{\sqrt{6}} 3 \right )^2 + \frac {1}{ \log_2 \sqrt{6} \cdot \log_{18} \sqrt{6} } = \ ?


The answer is 4.

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1 solution

Noel Lo
Mar 30, 2015

( l o g 6 3 ) 2 (log_{\sqrt{6}}{3})^2 + 1 l o g 2 6 l o g 18 6 \frac{1}{log_{2} \sqrt{6} * log_{18} \sqrt{6}}

= ( l o g 6 3 ) 2 (log_{\sqrt{6}}{3})^2 + log 6 2 \log_{\sqrt6}{2} * log 6 18 \log_{\sqrt{6}}{18}

= ( l o g 6 3 ) 2 (log_{\sqrt{6}}{3})^2 + log 6 ( 6 3 ) \log_{\sqrt6}(\frac{6}{3}) * log 6 ( 6 × 3 ) \log_{\sqrt{6}}(6 \times3)

= ( l o g 6 3 ) 2 (log_{\sqrt{6}}{3})^2 + ( log 6 6 l o g 6 3 \log_{\sqrt6}{6} - log_{\sqrt6}{3} ) ( log 6 6 + l o g 6 3 \log_{\sqrt6}{6} + log_{\sqrt6}{3} )

= ( l o g 6 3 ) 2 (log_{\sqrt{6}}{3})^2 + ( 2 l o g 6 3 (2- log_{\sqrt6}{3} ) ( 2 + l o g 6 3 2 + log_{\sqrt6}{3} )

= ( l o g 6 3 ) 2 (log_{\sqrt{6}}{3})^2 + 2 2 2^2 - ( l o g 6 3 ) 2 (log_{\sqrt{6}}{3})^2 = 4 \boxed{4}

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