Question

If $x={\log }_a\left(bc\right),y={\log }_b\left(ca\right)$ and $z={\log }_c\left(ab\right)$ , then which of the following equation is equal to $1$?

• A. $x+y+z$
• B. $(1+x{)}^{-1}+(1+y{)}^{-1}+(1+z{)}^{-1}$
• C. $xyz$
• D. none of these

Answer 1

Answer: (B)

$(1+x{)}^{-1}+(1+y{)}^{-1}+(1+z{)}^{-1}$

$x={\log }_{a}bc$
$\Rightarrow 1+x=\frac{\log b+\log c}{\log a}+1\Rightarrow 1+x=\frac{\log b+\log c+\log a}{\log a}$ ...(1)
$y={\log }_{b}ac$
$\Rightarrow 1+y=\frac{\log a+\log c}{\log b}+1\Rightarrow 1+y==\frac{\log a+\log c+\log b}{\log b}$ ...(2)
$z={\log }_{c}ab\Rightarrow 1+z=\frac{\log a+\log b}{\log c}+1$
$\Rightarrow 1+z=\frac{\log a+\log b+\log c}{\log c}$ ...(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: B