Question

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

Answer 1

To find $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$

Putting values we get $\frac{1}{{\log }_{a}bc+1}+\frac{1}{{\log }_{b}ca+1}+\frac{1}{{\log }_{c}ab+1}$

$=\frac{1}{\frac{\log bc}{\log a}+1}+\frac{1}{\frac{\log ca}{\log b}+1}+\frac{1}{\frac{\log ab}{\log c}+1}$

$=\frac{\log a}{\log bc+\log a}+\frac{\log b}{\log ca+\log b}+\frac{\log c}{\log ab+\log c}$

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

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