If $x = {log}_{2 a} a, y = {log}_{3 a} 2 a a n d z = {log}_{4 a} 3 a$ , then $x y z + 1 =$

2 y z

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2 x y

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2 z x

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n o n e

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Solution

The correct option is B $2 y z$
$x y z = \frac{log a}{log 2 a} . \frac{log 2 a}{log 3 a} . \frac{log 3 a}{log 4 a} = \frac{log a}{log 4 a}$
$x y z + 1 = \frac{log {4 a}^{2}}{log 4 a} = 2. \frac{log 2 a}{log 4 a} = 2. \frac{log 2 a}{log 3 a} . \frac{log 3 a}{log 4 a}$
$x y z + 1 = 2 y z$

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