TITLE: If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
QUESTION [5 upvotes]: If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .

My Working:
$\frac{\log a}{b-c}= \frac{\log b}{c-a}$
$ (c-a)\log a=(b-c) \log b$
$ \log a^{c-a}=\log b^{b-c}$
$ \frac {a^c}{a^a}=\frac{b^b}{b^c}$
$ \frac {a^c \cdot{b^c}}{a^a} =b^b\qquad \text{(i)}$
Similarly, taking the next two terms we obtain,
$b^b=\frac{b^a \cdot c^a}{c^c}\qquad \text{(ii)}$
I tried to solve the two equations obtained to get to the desired statement but I couldn't. Is the way adopted correct or is there another way to reach the desired answer. Please help me proceed with this question

REPLY [4 votes]: Let $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=t\ $ implies $a=e^{(b-c)t}$, $b=e^{(c-a)t}$ & $c=e^{(a-b)t}$
so now, using values of a, b, c: $$a^ab^bc^c=e^{a(b-c)t}e^{b(c-a)t}e^{c(a-b)t}=e^{(ab-ac+bc-ab+ac-bc)t}=e^0=1$$