Question

If $\frac{loga}{b-c}=\frac{logb}{c-a}=\frac{logc}{a-b}$, then evaluate $a^a.b^b.c^c$.

Answer 1

Let each ratio be k and $y=a^a{b}^b{c}^c$
Then,
$loga=k(b-c)$
$logb=k(c-a)$
$logc=k(a-b)$
and
$logy=a^a{b}^b{c}^c$
$logy=aloga+blogb+clogc$
$logy=k(b-c)+k(c-a)+k(a-b)$
$logy=k(b-c+c-a+a-b)$
$\therefore logy=0\Rightarrow y=1$

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