Question

If $x = \log_a bc, y = \log_b ca, z = \log_c ab,$ then $\frac{x}{1+x} + \frac{y}{1+y} + \frac{z}{1+z } =$

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

$x=\log _{a} bc , y = \log _{b} ca ,z =\log _{c} ab$
$\Rightarrow x =\frac{\log bc}{\log a} , y = \frac{\log ca}{\log b} , z = \frac{\log ab}{\log c}$
$\therefore \, \frac{x}{1+x} + \frac{y}{1+y} + \frac{z}{1+z}$
$= \frac{\log bc}{\log a + \log bc} + \frac{\log ca}{\log b+\log ca} + \frac{\log ab}{\log c +\log ab}$
$= \frac{\log b+\log c}{\log a +\log b +\log c} + \frac{\log c+\log a}{\log b+\log c+\log a} + \frac{\log a +\log b}{\log c+\log a+\log b}$
$= \frac{2\left(\log a+\log b+\log c\right)}{\log a+\log b+\log c} = 2$