Question

Solve the following: $\frac{1}{{\log }_{xy}xyz}+\frac{1}{{\log }_{yz}xyz}+\frac{1}{{\log }_{zx}xyz}=$

• A. 0
• B. 1
• C. 2
• D. $lo{g}_{x}xyz$

Answer 1

The correct option is C 2

We have,

$\frac{1}{{\log }_{xy}xyz}+\frac{1}{{\log }_{yz}xyz}+\frac{1}{{\log }_{zx}xyz}$

$\Rightarrow \frac{{\log }_{xy}xy}{{\log }_{xy}xyz}+\frac{{\log }_{yz}yz}{{\log }_{yz}xyz}+\frac{{\log }_{zx}zx}{{\log }_{zx}xyz}$

We know that

$\frac{{\log }_{a}m}{{\log }_{a}n}={\log }_{n}m$

Therefore,

$\Rightarrow {\log }_{xyz}xy+{\log }_{xyz}yz+{\log }_{xyz}zx$

$\Rightarrow {\log }_{xyz}\left(x^2{y}^2{z}^2\right)$

$\Rightarrow {\log }_{xyz}(xyz{)}^2$

$\Rightarrow 2{\log }_{xyz}\left(xyz\right)$

$\Rightarrow 2$

Hence, this is the answer.