Question

If $x=1+{\log }_{a}bc,y=1+{\log }_{b}ca,z=1+{\log }_{c}ab$, then the value of $xyz$ will be

• A. $xy+yz+zx$
• B. $x+y+z$
• C. $x-y-z$
• D. $xy-yz-zx$

Answer 1

The correct option is A $xy+yz+zx$

Since $x-1={\log }_{a}bc,y-1={\log }_{b}ca$ and $z-1={\log }_{c}ab$

$x-1={\log }_a\left(\frac{abc}{a}\right)={\log }_a\left(abc\right)-{\log }_{a}a={\log }_a\left(abc\right)-1,$

$\left[\log \right(x/y)=\log x-\log y]and{\log }_{a}a=1]$
$\therefore x={\log }_a\left(abc\right)\Rightarrow \frac{1}{x}={\log }_{abc}a$

Similarly $\frac{1}{y}={\log }_{abc}b$ & $\frac{1}{z}={\log }_{abc}c$
$\therefore$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}={\log }_{abc}a+{\log }_{abc}b+{\log }_{abc}c={\log }_{abc}\left(abc\right)=1$
$\therefore xyz=xy+yz+zx$