Question

If $x=1+{\log }_{p}qr,y=1+{\log }_{q}rp,z=1+{\log }_{r}pq$ then the value of $(xy+yz+zx-xyz)$ is

• A. $1$
• B. $0$
• C. $-1$
• D. None

Answer 1

Answer: (B)

$0$

$x=1+{\log }_{p}qr={\log }_{p}p+{\log }_{p}qr={\log }_{p}pqr$

Similarly, $y={\log }_{q}pqr$ and $z={\log }_{r}pqr$

$\therefore xy+yz+zx-xyz$

$=\left({\log }_{p}pqr\right)-\left({\log }_{q}pqr\right)+\left({\log }_{q}pqr\right).\left({\log }_{r}pqr\right)+\left({\log }_{r}pqr\right)\left({\log }_{p}pqr\right)-\left({\log }_{p}pqr\right)\left({\log }_{q}pqr\right)\left({\log }_{r}pqr\right)$

$=[\frac{\log \left(pqr\right)}{\log p}+\frac{\log \left(pqr\right)}{\log q}+....]-\frac{\left(\log \right(pqr){)}^3}{\log p.\log q.\log r}$

$=\left[\log \right(pqr){]}^2[\frac{1}{\log p.\log q}+\frac{1}{\log q.\log r}+\frac{1}{\log r.\log p}]-[\log \left(pqr\right){]}^3\left[\frac{1}{\log p.\log q.\log r}\right]$

$=\left[\log \right(pqr){]}^2\left[\frac{\log p+\log q+\log r}{\log p.\log q.\log r}\right]-[\log \left(pqr\right){]}^3\left[\frac{1}{\log p.\log q.\log r}\right]$

$=\left[\log \right(pqr){]}^2\left[\frac{\log \left(pqr\right)}{\log p.\log q.\log r}\right]-[\log \left(pqr\right){]}^3\left[\frac{1}{\log p.\log q.\log r}\right]$

$=0$