Question

If $a=1+{\log }_{x}yz$, $b=1+{\log }_{y}zx$ and $c=1+{\log }_{z}xy$, then the value of $(ab+bc+ca)$ is equal to

• A. $1$
• B. $0$
• C. $abc$
• D. $z$

Answer 1

The correct option is C $abc$

$a=1+{\log }_{x}yz={\log }_{x}x+{\log }_{x}yz={\log }_{x}xyz$
Similarly $b={\log }_{y}xyz$
and $c={\log }_{z}xyz$
Now, $ab+bc+ca=abc[\frac{1}{c}+\frac{1}{a}+\frac{1}{b}]$

$=abc[\frac{1}{{\log }_{x}xyz}+\frac{1}{{\log }_{y}xyz}+\frac{1}{{\log }_{z}xyz}]$

$=abc[{\log }_{xyz}x+{\log }_{xyz}y+{\log }_{xyz}z]$

$=abc\left[{\log }_{xyz}xyz\right]=abc$