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+1 vote
9.8k views
in Logarithm by (24.0k points)
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If x = log2a a, y = log3a2a, z = log4a3a, then xyz – 2yz equals

(a) a3 

(b) 1 

(c) 0 

(d) –1

1 Answer

+2 votes
by (23.5k points)
selected by
 
Best 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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