Unable to solve logarithm question

Question

Given

$$\dfrac{a(b+c-a)}{\log a}=\dfrac{b(c+a-b)}{\log b}=\dfrac{c(a+b-c)}{\log c}$$

To prove:

$$a^bb^a=b^cc^b=c^aa^c$$

What i tried is

$$\log (a^z)=a(b+c-a)$$ and similarly for other two. I am unable to break this down further! please help!

Answer 1

Just like in your other question, $$\log a = \lambda\cdot a(b+c-a)$$ and so on give: $$ b\log a +a\log b = \lambda\cdot\left(ab(b+c-a)+ab(a+c-b)\right)=2\lambda\cdot abc$$ that is symmetric in $a,b,c$, so by exponentiating the previous line $$ a^b b^a = a^c c^a = b^c c^b $$ follows.