Question

If $a^p=x,a^q=y,x^q{y}^p=a^{2/r}$ then $pqr$ is:

• A. $1$
• B. $-1$
• C. $x^{-1/p}$
• D. $x^p$

Answer 1

The correct option is A $1$

$a^p=x\Rightarrow {\log }_a{a}^p={\log }_{a}x$

$\Rightarrow p={\log }_{a}x$

Similarly, $q={\log }_{a}y$

$x^q{y}^p=a^{2/r}$

${\log }_a(x^q.y^p)={\log }_a\left(a^{2/r}\right)$

$\Rightarrow \left({\log }_a{x}^q\right)+\left({\log }_a{y}^p\right)=\frac{2}{r}.{\log }_a\left(a\right)$

$\Rightarrow q.{\log }_{a}x+p.{\log }_{a}y=\frac{2}{r}$

$\Rightarrow q.p+p.q=\frac{2}{r}$

$\Rightarrow 2pq=\frac{2}{r}$

$\Rightarrow pq=\frac{1}{r}$

$\Rightarrow pqr=1$