Question

The value of the expression
$\frac{1}{1+{\log }_{b}a+{\log }_{b}c}+\frac{1}{1+{\log }_{c}a+{\log }_{c}b}+\frac{1}{1+{\log }_{a}b+{\log }_{a}c}$ is equal to

• A. $abc$
• B. $\frac{1}{abc}$
• C. $0$
• D. $1$

Answer 1

Answer: (D)

$1$

Consider, $\frac{1}{1+{\log }_{b}a+{\log }_{b}c}+\frac{1}{1+{\log }_{c}a+{\log }_{c}b}+\frac{1}{1+{\log }_{a}b+{\log }_{a}c}$

$=\frac{1}{1+\frac{\log a}{\log b}+\frac{\log c}{\log b}}+\frac{1}{1+\frac{\log a}{\log c}+\frac{\log b}{\log c}}+\frac{1}{1+\frac{\log b}{\log a}+\frac{\log c}{\log a}}$ $[\because {\log }_{y}x=\frac{\log x}{\log y}]$

$=\frac{\log b}{\log b+\log a+\log c}+\frac{\log c}{\log c+\log a+\log b}+\frac{\log a}{\log c+\log a+\log b}=1$.