Question

Find the value of $\frac{1}{1+{\log }_{b}a+{\log }_{b}c}+\frac{1}{1+{\log }_{c}a+{\log }_{c}b}+\frac{1}{1+{\log }_{a}b+{\log }_{a}c}$.

• A. $2$
• B. $1$
• C. $0$
• D. none of these

Answer 1

Answer: (B)

$1$

Now,

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

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

$=\frac{1}{{\log }_{b}abc}+\frac{1}{{\log }_{c}abc}+\frac{1}{{\log }_{a}abc}$ [ Using product rule of logarithtm]

$=\frac{\log b}{\log abc}+\frac{\log c}{\log abc}+\frac{\log a}{\log abc}$

$=\frac{\log abc}{\log abc}$

$=1$.