If $\frac {\ln (x)}{(b-c)}=\frac {\ln (y)}{(c-a)}=\frac {\ln (z)}{(a-b)}$

Question

If $\dfrac {\ln (x)}{(b-c)}=\dfrac {\ln (y)}{(c-a)}=\dfrac {\ln (z)}{(a-b)}$ then prove that: $(yz)^{a}.(zx)^{b}.(xy)^{c}=1$

My Attempt: Given, $$\dfrac {\ln (x)}{(b-c)}=\dfrac {\ln (y)}{(c-a)}=\dfrac {\ln (z)}{(a-b)}=k(\textrm {let})$$ so we get: $$\ln (x)=k(b-c)$$ $$\ln (y)=k(c-a)$$ $$\ln (z)=k(a-b)$$ Now, what to do next?

Answer 1

Then $x = e^{k(b-c)}$ , $ y= e^{k(c-a)}$ and $z= e^{k(a-b)}$.

So $x^{b+c} = e^{k(b^2-c^2)}$ , $ y^{a+c}= e^{k(c^2-a^2)}$ and $z^{a+b}= e^{k(a^2-b^2)}$.

and finally: \begin{eqnarray}(yz)^{a}.(zx)^{b}.(xy)^{c} &=& x^{b+c} \cdot y^{a+c}\cdot z^{a+b}\\ &=&e^{k(b^2-c^2)}\cdot e^{k(c^2-a^2)}\cdot e^{k(a^2-b^2)} \\ &=& e^{k(a^2-b^2+b^2-c^2+c^2-a^2)} \\ &=& e^0 = 1 \end{eqnarray}

Answer 2

$$a(\ln y+\ln z)+b(\ln z+\ln x)+c(\ln x+\ln y)=k\sum_{\rm cyc}a(c-b)=0.$$

Answer 3

Note that

$$\begin{cases}\ln x + \ln y= \ln xy = -\ln z\\\\\ln y + \ln z= \ln yz = -\ln x\\\\\ln z + \ln x= \ln zx = -\ln y\end{cases}$$

thus

$$c \ln xy + a \ln yz + b \ln zx = \ln [(xy)^c\cdot(yz)^a\cdot(zx)^b]=$$

$$=- c\ln z-a\ln x-b\ln y=-kc(a-b)-ka(b-c)-kb(c-a)=0$$

and therefore

$$(xy)^c\cdot(yz)^a\cdot(zx)^b=1$$