Question

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

Answer: (B)

$1$

Now, $1+x={\log }_{a}a+{\log }_{a}bc={\log }_{a}abc$
$\Rightarrow \frac{1}{1+x}={\log }_{abc}a$
Similarly, $\frac{1}{1+y}={\log }_{abc}b$ and $\frac{1}{1+z}={\log }_{abc}c$
$\therefore \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$
$={\log }_{abc}a+{\log }_{abc}b+{\log }_{abc}c$
$={\log }_{abc}abc=1$