Question

The value of $\frac{1}{{\log }_{xy}xyz}+\frac{1}{{\log }_{yz}xyz}+\frac{1}{{\log }_{zx}xyz}$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. ${\log }_{x}xyz$

Answer 1

Answer: (C)

$2$

Let, $a=\frac{1}{{\log }_{xy}xyz}+\frac{1}{{\log }_{yz}xyz}+\frac{1}{{\log }_{zx}xyz}$
$\Rightarrow a={\log }_{xyz}xy+{\log }_{xyz}yz+{\log }_{xyz}zx[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
$\Rightarrow a={\log }_{xyz}(xy\cdot yz\cdot zx)[\because \log a+\log b+\log c=\log (abc\left)\right]$
$\Rightarrow a={\log }_{xyz}(xyz{)}^2=2{\log }_{x}yz(xyz)=2[\because {\log }_{a}a=1]$
Ans: C