Q. If $a,\ b,\ c$ are distinct positive numbers each being different from 1 such that $\left( \log_{b}{a.\log_{c}{a - \log_{a}a}} \right) + \left( \log_{a}{b.\log_{c}{b - \log_{b}b}} \right) + \left( \log_{a}{c.\log_{b}{c - \log_{c}c}} \right) = 0$, then $abc$ is

A) 0
B) $e$
C) 1
D) 2
Explanation

(c)
$\left( \log_{b}{a.\log_{c}{a - \log_{a}a}} \right) + (\log_{a}{b.\log_{c}{b - \log_{b}b}} + \left( \log_{a}{c.\log_{b}{c - \log_{c}c}} \right) = 0$
$\Rightarrow \ \left( \frac{\log a}{\log b}.\ \frac{\log a}{\log c} - \frac{\log a}{\log a} \right) + \left( \frac{\log b}{\log a}.\frac{\log b}{\log c} - \frac{\log b}{\log b} \right) + \left( \frac{\log c}{\log a}.\frac{\log c}{\log b} - \frac{\log c}{\log c} \right) = 0$
$\Rightarrow \left( \log a \right)^{3} + \left( \log b \right)^{3} + \left( \log c \right)^{3} - 3\log{a\log{b\log{c = 0}}}$
$\Rightarrow \ \ \ \left( \log{a + \log{b + \log c}} \right) = 0$
$\left( \begin{array}{r}
\because if\ a^{3} + b^{3} + c^{3} = 3abc, \\
then\ a + b + c = 0
\end{array} \right)$
$\Rightarrow abc = 1$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.