If $K = \frac{\log_{2}^{24}}{\log_{96}^{2}} - \frac{\log_{2}^{192}}{\log_{12}^{2}},$ then the values of $\log_{81}^{K^{100}}$ is_____

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100

$\frac{25}{2}$

answer is B.

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Detailed Solution

$K = \frac{\log_{2}^{24}}{\log_{96}^{2}} - \frac{\log_{2}^{192}}{\log_{12}^{2}} = (\log_{2} 24) (\log_{2} 96) - (\log_{2} 192) (\log_{2} 12) = (\log_{2} 2 + \log_{2} 12) (\log_{2} 2^{3} + \log_{2} 12) - (\log_{2} 12) (\log_{2} 2^{4} + \log_{2} 12) = (1 + \log_{2} 12) (3 + \log_{2} 12) - (\log_{2} 12) (4 + \log_{2} 12) = 3 + \log_{2} 12 + 3 \log_{2} 12 + \log_{2} 12 \cdot \log_{2} 12 - (4 \log_{2} 12 + \log_{2} 12 \cdot \log_{2} 12) = 3$
$\log_{81}^{K^{100}} = \log {}_{3^{4}}3^{100} = \frac{100}{4} \log {}_{3}3 = 25$