Question

If a,b,c are distinct positive numbers,each different from 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 $abc=$

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is A $1$

We have,

$[{\log }_{b}a{\log }_{c}a-{\log }_{a}a]+[{\log }_{a}b{\log }_{c}b-{\log }_{b}b]+[{\log }_{a}c{\log }_{b}c-{\log }_{c}c]=0$

$[{\log }_{b}a{\log }_{c}a-1]+[{\log }_{a}b{\log }_{c}b-1]+[{\log }_{a}c{\log }_{b}c-1]=0$

${\log }_{b}a{\log }_{c}a+{\log }_{a}b{\log }_{c}b+{\log }_{a}c{\log }_{b}c-3=0$

${\log }_{b}a{\log }_{c}a+{\log }_{a}b{\log }_{c}b+{\log }_{a}c{\log }_{b}c=3$

$\frac{\log a}{\log b}\times \frac{\log a}{\log c}+\frac{\log b}{\log a}\times \frac{\log b}{\log c}+\frac{\log c}{\log a}\times \frac{\log c}{\log b}=3$

$\frac{(\log a{)}^2}{\log b\log c}+\frac{(\log b{)}^2}{\log a\log c}+\frac{(\log c{)}^2}{\log a\log b}=3$

$(\log a{)}^3+(\log b{)}^3+(\log c{)}^3=3(\log a\log b\log c)$

$(\log a{)}^3+(\log b{)}^3+(\log c{)}^3-3(\log a\log b\log c)=0$

We know that

$(x^3+y^3+z^3)-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Therefore,

$(\log a+\log b+\log c)\left(\right(\log a{)}^2+(\log b{)}^2+(\log c{)}^2-\log a\log b-\log b\log c-\log c\log a)=0$

$(\log a+\log b+\log c)=0$

$\log abc=0$

$abc=e^0$

$abc=1$

Hence, this is the answer.