Question

If $a,b,c>1$, $\Delta =\left|\begin{array}{ccc}{\log }_a\left(abc\right)& {\log }_{a}b& {\log }_{a}c\\ {\log }_b\left(abc\right)& 1& {\log }_{b}c\\ {\log }_c\left(abc\right)& {\log }_{c}b& 1\end{array}\right|$ is

• A. $0$
• B. ${\log }_{a}b+{\log }_{b}c+{\log }_{c}a$
• C. ${\log }_{abc}(a+b+c)$
• D. none of these

Answer 1

The correct option is A $0$

$\Delta =\left|\begin{array}{ccc}{\log }_a\left(abc\right)& {\log }_{a}b& {\log }_{a}c\\ {\log }_b\left(abc\right)& 1& {\log }_{b}c\\ {\log }_c\left(abc\right)& {\log }_{c}b& 1\end{array}\right|$

$={\log }_a\left(abc\right)(1-{\log }_{c}b{\log }_{b}c)-{\log }_b\left(abc\right)({\log }_{a}b-{\log }_{c}b{\log }_{a}c)+{\log }_c\left(abc\right)({\log }_{a}b{\log }_{b}c-{\log }_{a}c)$

$=\frac{logabc}{logb}(\frac{logb}{loga}-\frac{logb}{logc}\frac{logc}{loga})+\frac{logabc}{logc}(\frac{logb}{loga}\frac{logc}{logb}-\frac{logc}{loga})=0$

Hence, option 'A' is correct.