Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is A $1$

$x={\log }_{a}bc$
$\Rightarrow 1+x=\frac{\log b+\log c}{\log a}+1\Rightarrow 1+x=\frac{\log b+\log c+\log a}{\log a}[\because {\log }_{a}b=\frac{\log b}{\log a}\text{\&}\log (ab)=\log a+\log b]$ ...(1)
similarly,
$y={\log }_{b}ac$
$\Rightarrow 1+y=\frac{\log a+\log c}{\log b}+1\Rightarrow 1+y=\frac{\log a+\log c+\log b}{\log b}$ ...(2)
$z={\log }_{c}ab\Rightarrow 1+z=\frac{\log a+\log b}{\log c}+1$
$\Rightarrow 1+z=\frac{\log a+\log b+\log c}{\log b}$ ...(3)
From (1), (2) & (3), we get
${(1+x)}^{-1}+{(1+y)}^{-1}+{(1+z)}^{-1}=\frac{\log a+\log b+\log c}{\log a+\log b+\log c}=1$
Ans: A