Question

If $a=lo{g}_x\left(yz\right),b=lo{g}_y\left(zx\right),c=lo{g}_z\left(xy\right)$ where $x,y,z$ are positive reals not equal to unity then $abc-a-b-c$ is equal to -

• A. $2$
• B. $1$
• C. $-1$
• D. $0$

Answer 1

The correct option is A $2$

$a+1=lo{g}_x\left(x\right)+lo{g}_x\left(yz\right)=lo{g}_x\left(xyz\right)$
$b+1=lo{g}_y\left(xyz\right)$
$c+1=lo{g}_z\left(xyz\right)$
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=lo{g}_{xyz}\left(x\right)+lo{g}_{xyz}\left(y\right)+lo{g}_{xyz}\left(z\right)$
$\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1$
$\Rightarrow (a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)=(a+1)(b+1)(c+1)$
$\Rightarrow ab+a+b+1+bc+b+c+1+ac+a+c+1=abc+1+ab+bc+ca+a+b+c$
$\Rightarrow a+b+c-abc=-2$
$\Rightarrow abc-a-b-c=2$