InterviewSolution
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InterviewSolution
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If x = log2a a, y = log3a2a, z = log4a3a, then xyz – 2yz equals(a) a3 (b) 1 (c) 0 (d) –1 |
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Answer» (d) -1 \(x\) = log2a a = \(\frac{\text{log}\,a}{\text{log}\,2a}\), y = log3a 2a = \(\frac{\text{log}\,2a}{\text{log}\,3a}\) z = log4a 3a = \(\frac{\text{log}\,3a}{\text{log}\,4a}\) ∴ xyz - 2yz = \(\frac{\text{log}\,a}{\text{log}\,2a}\).\(\frac{\text{log}\,2a}{\text{log}\,3a}\).\(\frac{\text{log}\,3a}{\text{log}\,4a}\) - 2\(\frac{\text{log}\,2a}{\text{log}\,3a}\).\(\frac{\text{log}\,3a}{\text{log}\,4a}\) = \(\frac{\text{log}\,a}{\text{log}\,4a}\) - 2\(\frac{\text{log}\,2a}{\text{log}\,4a}\) = \(\frac{\text{log}\,a-2\,\text{log}\,2a}{\text{log}\,4a}\) = \(\frac{\text{log}\,a-\,\text{log}\,(2a^2)}{\text{log}\,4a}\) = \(\frac{\text{log}\frac{a}{4}a^2}{\text{log}\,4a}\) = \(\frac{\text{log}\,(4a)^{-1}}{\text{log}\,(4a)}\) = \(\frac{-1.\text{log}\,4a}{\text{log}\,4a}\) = -1. |
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