If $p = {log}_{2 a} a, q = {log}_{3 a} 2 a$ and $r = {log}_{4 a} 3 a$ , then find the value of $q r (2 - p)$ .

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Solution

$q r = {log}_{3 a} 2 a {log}_{4 a} 3 a = {log}_{4 a} 2 a$
$p q r = {log}_{2 a} a {log}_{4 a} 2 a = {log}_{4 a} a$
Now,
$q r (2 - p) = 2 q r - p q r$
$= 2 {log}_{4 a} 2 a - {log}_{4 a} a$
$= {log}_{4 a} 4 a^{2} - {log}_{4 a} a$
$= {log}_{4 a} \frac{4 a^{2}}{a} = {log}_{4 a} 4 a = 1$

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