Question

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

• A. $x+y+z$
• B. $1$
• C. $ab+bc+ca$
• D. $abc$

Answer 1

The correct option is B $1$

Given : $x={\log }_{a}bc$, $y={\log }_{b}ca$ and $z={\log }_{c}ab$

Here, $1+x={\log }_{a}a+{\log }_{a}bc={\log }_{a}abc$

$\Rightarrow \frac{1}{1+x}={\log }_{abc}a$

Similarly, $\frac{1}{1+y}={\log }_{abc}b$
and $\frac{1}{1+z}={\log }_{abc}c$

$\therefore \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$

$={\log }_{abc}a+{\log }_{abc}b+{\log }_{abc}c$

$={\log }_{abc}abc=1$