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$ ...(i)

Similarly, taking the next two terms we obtain,

$b^b=\frac{b^a \cdot c^a}{c^c}$...(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

  • 1
    $\begingroup$ Don't we even get $a=b=c=1$? From your calculation above follows $a^a b^b c^c = (abc)^a = (abc)^b = (abc)^c$, from which we get $a=b=c$ and then $a^{3a} =1$ gives us $a=0$ or $a=1$. Maybe it doesn't hold for complex values though... $\endgroup$ – Hyperplane Dec 12 '16 at 11:39

It is given that $$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}\tag1$$

Now $$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log a+\log b}{b-c+c-a} \,\,\,\,\,\,\,\text {(by Addendo)}$$ $$=\frac{\log ab}{b-a} \tag2$$

So from $(1)$ and $(2)$, we get that $$\frac{\log c}{a-b}=\frac{\log ab}{b-a}$$ $$\implies ab =\frac {1}{c}\tag3$$

From equation $(i)$ established by you in the question and $(3) $, we get $$\frac {a^c \cdot{b^c}}{a^a} =b^b$$ $$\implies a^c \cdot b^c=a^a \cdot b^b$$ $$\implies \frac {1}{c^c}=a^a \cdot b^b$$ $$\implies a^a \cdot b^b \cdot c^c= 1$$

Hope this helps you.


Call $k$ the common value of $$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$$ and then $$\log(a^ab^bc^c)=a\log a+b\log b+c\log c=\left(a(b-c)+b(c-a)+c(a-b)\right)\cdot k=0$$ which implies that $a^ab^bc^c=1$.

  • $\begingroup$ Thanks, could you also add how I could proceed with my method as I wasn't able to get to the answer using my method $\endgroup$ – Osheen Sachdev Dec 12 '16 at 11:16

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: $$a^ab^bc^c=e^{a(b-c)t}e^{b(c-a)t}e^{c(a-b)t}=e^{(ab-ac+bc-ab+ac-bc)t}=e^0=1$$


We use following property if $F=\frac{A_1}{B_1}=\frac{A_2}{B_2}$ then $$F=\frac{A_1+A_2}{B_1+B_2}.$$

So, your equality implies





  • $\begingroup$ Can you elaborate on your second step $\endgroup$ – Osheen Sachdev Dec 12 '16 at 11:23
  • $\begingroup$ Perfect, Thanks a lot! $\endgroup$ – Osheen Sachdev Dec 12 '16 at 11:31
  • $\begingroup$ This is pretty much the same as the answer of Robert Z, but poorly explained (division by $0$!). $\endgroup$ – Arnaud D. Aug 27 '18 at 15:04
  • $\begingroup$ @ArnaudD. Please look again. i edited. $\endgroup$ – hamam_Abdallah Sep 1 '18 at 20:37

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