Question

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

$1$

Explanation for the correct answer:

Find the value of $\frac{1}{(1+x)}+\frac{1}{(1+y)}+\frac{1}{(1+z)}$:

Given,

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

Now, add $1$ on both sides in each function.

$\begin{array}{rcl}1+x& =& 1+{\log }_{a}bc\\ & =& {\log }_{a}a+{\log }_{a}bc\left[\because {\log }_{a}a=1\right]\\ & =& {\log }_{a}abc\left[\because \log a+\log b=\log ab\right]\\ 1+y& =& 1+{\log }_{b}ca\\ & =& {\log }_{b}b+{\log }_{b}ca\\ & =& {\log }_{b}abc\\ 1+z& =& 1+{\log }_{c}ab\\ & =& {\log }_{c}c+{\log }_{c}ab\\ & =& {\log }_{c}abc\end{array}$

So,

$\begin{array}{rcl}\frac{1}{(1+x)}+\frac{1}{(1+y)}+\frac{1}{(1+z)}& =& \frac{1}{{\log }_{a}abc}+\frac{1}{{\log }_{b}abc}+\frac{1}{{\log }_{c}abc}\\ & =& {\log }_{abc}a+{\log }_{abc}b+{\log }_{abc}c\left[\because \frac{1}{{\log }_{a}b}={\log }_{b}a\right]\\ & =& {\log }_{abc}abc\\ & =& 1\left[\because {\log }_{x}x=1\right]\end{array}$

Hence, the correct option is B.