Question

${\log }_{a}bc=x,{\log }_{b}ac=y,{\log }_{c}ab=z$, then $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=$

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. none

Answer 1

The correct option is B $1$

We have,

$x={\log }_{a}bc$

$x=\frac{\log bc}{\log a}$ $[\because {\log }_{a}m=\frac{\log m}{\log a}]$

Similarly,

$y=\frac{\log ac}{\log b}$

$z=\frac{\log ab}{\log c}$

Consider, $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$

$\Rightarrow \frac{1}{\frac{\log bc}{\log a}+1}+\frac{1}{\frac{\log ac}{\log b}+1}+\frac{1}{\frac{\log ab}{\log c}+1}$

$\Rightarrow \frac{\log a}{\log bc+\log a}+\frac{\log b}{\log ac+\log b}+\frac{\log c}{\log ab+\log c}$

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

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

$\Rightarrow \frac{\log abc}{\log abc}$

$\Rightarrow 1$

Hence, this is the answer.