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Property 3

If x=log 2a...

Question

If $x = {log}_{2 a} a, y = {log}_{3 a} 2 a, z = {log}_{4 a} 3 a$ , prove that $1 + x y z = 2 y z$ .

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Solution

$L H S = 1 + x y z$
$= 1 + ({log}_{2 a} a) ({log}_{3 a} 2 a) ({log}_{4 a} 3 a)$

$= 1 + \frac{log a}{log 2 a} \frac{log 2 a}{log 3 a} \frac{log 3 a}{log 4 a} \dots \dots {∵ {log}_{b} a = \frac{log a}{log b}}$

$= 1 + \frac{log a}{log 4 a}$
$= {log}_{4 a} 4 a + {log}_{4 a} a \dots \dots {∵ {log}_{a} a = 1}$
$= {log}_{4 a} 4 a^{2} \dots \dots {∵ log a b = log a + log b}$
$= 2 {log}_{4 a} 2 a \dots \dots {∵ log a^{b} = b log a}$
$= 2 \frac{log 2 a}{log 4 a} = 2 \frac{log 2 a \times log 3 a}{log 3 a \times log 4 a}$
$= 2 ({log}_{3 a} 2 a) ({log}_{4 a} 3 a) = 2 y z = R H S$

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