Question

The value of the expression $\frac{1}{{\log }_{\sqrt{bc}}abc}+\frac{1}{{\log }_{\sqrt{ca}}abc}+\frac{1}{{\log }_{\sqrt{ab}}abc}$ is equal to

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B $1$

$\frac{1}{{\log }_{\sqrt{bc}}abc}+\frac{1}{{\log }_{\sqrt{ca}}abc}+\frac{1}{{\log }_{\sqrt{ab}}abc}$

$=\frac{\log \sqrt{bc}}{\log \left(abc\right)}+\frac{\log \sqrt{ca}}{\log \left(abc\right)}+\frac{\log \sqrt{ab}}{\log \left(abc\right)}[\because {\log }_{a}b=\frac{\log b}{\log a}]$

$=\frac{\frac{1}{2}(\log b+\log c)}{\log a+\log b+\log c}+\frac{\frac{1}{2}(\log a+\log c)}{\log a+\log b+\log c}+\frac{\frac{1}{2}(\log a+\log b)}{\log a+\log b+\log c}=1$

$[\because \log a^m=m\log a\text{\&}\log (xyz)=\log x+\log y+\log z]$
Ans: B