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If $x = \log_{2 a}^{a} : y = \log_{3 a}^{2 a} : z = \log_{4 a}^{3 a}$ then $(2 - x) yz =$

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Detailed Solution

$xyz = \log_{4 a}^{a} \Rightarrow 1 + xyz = \log_{4 a} 4 a + \log_{4 a} a = \log_{4 a}^{4 a^{2}} 2 y z = 2 \log_{3 a} 2 a \log_{4 a} 3 a = 2 \log_{4 a} 2 a = \log_{4 a} 4 a^{2}$

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