Question

If $\frac{loga}{b-c}=\frac{logb}{c-a}=\frac{logc}{a-b}$, then find the value of $a^a.b^b.c^c$

• A. 1

Answer 1

The correct option is A 1
Let $\frac{loga}{b-c}=\frac{logb}{c-a}=\frac{logc}{a-b}=k$

and $y=a^a{b}^b{c}^c$

$\Rightarrow$ $loga=k(b-c)$
$logb=k(c-a)$
$logc=k(a-b)$

and $logy=log\left(a^a{b}^b{c}^c\right)$

$logy=aloga+blogb+clogc$
$logy=ak(b-c)+bk(c-a)+ck(a-b)$
$logy=k(ab-ac+bc-ba+ca-cb)$

$\therefore logy=0\Rightarrow y=1$

or $a^a{b}^b{c}^c=1$