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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!

Max Payne
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1 Answers1

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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.

Community
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Jack D'Aurizio
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    How do you find methods of proceeding? thank yo again very much! – Max Payne May 01 '15 at 11:34
  • Please tell how did you know how to proceed? – Max Payne May 01 '15 at 11:59
  • @TimKrul: since you have to prove that three quantities are the same, and they are given by cyclic shifts of the variables, the key is just to find a symmetric expression in $a,b,c$. Moreover, there are not so many ways to go from the first line to the last one. – Jack D'Aurizio May 01 '15 at 12:27