Q. If $x = \log_{a}{bc,\ y = \log_{b}{ca,\ z = \log_{a}{ab}}}$, then the value of $\frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z}$ will be
A) $x + y + z$
B) 1
C) $ab + bc + ca$
D) $abc$
Explanation
(b)
Now,$1 + x = \log_{a}{a + \log_{a}{bc = \log_{a}{abc}}}$
$\Rightarrow \ \frac{1}{1 + x} = \log_{abc}a$
Similarly, $\frac{1}{1 + y} = \log_{abc}b\$ and $\frac{1}{1 + z} = \log_{abc}c$
$\therefore\ \frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z}$
$= \log_{abc}{a + \log_{abc}{b + \log_{abc}c}}$
$= \log_{abc}{abc = 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.