Question

If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ then prove that $a^a.b^b.c^c=1$

Answer 1

Given $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}\text{Let}\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k$
Consider $\frac{\log a}{b-c}=k$
$\log a=k(b-c)$
$\therefore a=e^{(b-c)k}$ (Since, $\log a=x\Rightarrow a=e^x$)
Similar way, $\frac{\log b}{c-a}=k$
$\log b=k(c-a)$
$\therefore b=e^{(c-a)k}$
$\frac{\log c}{a-b}=k$
$\log c=k(a-b)$
$\therefore c=e^{(a-b)k}$
Consider $a^a\cdot b^b\cdot c^c={\left(e^{(b-c)k}\right)}^a{\left(e^{(c-a)k}\right)}^b{\left(e^{(a-b)k}\right)}^c$
$=\left(e^{a(b-c)k}\right)\left(e^{b(c-a)k}\right)\left(e^{c(a-b)k}\right)$ (Since, $(a^m{)}^n=a^{(m\times n)}$)
$=e^{a(b-c)k+b(c-a)k+c(a-b)k}$
$=e^{(ab-ac+bc-ab+ac-bc)k}$
$=e^0$
$=1$ (Since, $a^0=1$ for any non zero quantity of $a$)
$\therefore a^a.b^b.c^c=1$