Question

If $a,b,c$ are distinct positive numbers not equal to $1$ such that $({\log }_{b}a{\log }_{c}a-{\log }_{a}a)+({\log }_{a}b{\log }_{c}b-{\log }_{b}b)+({\log }_{a}c{\log }_{b}c-{\log }_{c}c)=0$, then $abc$ is

• A. $0$
• B. $e$
• C. $1$
• D. none of these

Answer 1

The correct option is C $1$

We have,
$({\log }_{b}a\cdot {\log }_{c}a-{\log }_{a}a)+({\log }_{a}b\cdot {\log }_{c}b-{\log }_{b}b)+({\log }_{a}c\cdot {\log }_{b}c-{\log }_{c}c)=0$
Lets assume,
$\log a=x$
$\log b=y$
$\log c=z$

Substituting in the equation, we get
$(\frac{x}{y}.\frac{x}{z}-1)+(\frac{y}{x}\cdot \frac{y}{z}-1)+(\frac{z}{x}.\frac{z}{y}-1)=0,(\because {\log }_{q}p=\frac{\log p}{\log q}\text{\&}{\log }_{p}p=1)$
$\Rightarrow \frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{yx}=3$
$\Rightarrow x^3+y^3+z^3=3xyz$

This implies that
$(x+y+z{)}^3=0$
$\Rightarrow x+y+z=0$
$\Rightarrow \log a+\log b+\log c=0$
$\Rightarrow \log \left(abc\right)=\log \left(1\right)$
$\therefore abc=1$