<s>[INST] Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$ Given:
$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}$$
We have to show that :
$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$
I made three equations using cross multiplication :
$$1.~~x^{c-a}=y^{b-c}$$
$$2.~~y^{a-b}=z^{c-a}$$
$$3.~~z^{b-c}=x^{a-b}$$
How do I proceed hereafter? If I multiply the equations, one variable goes away from exponents. 
Thank you.
 [/INST] First let's assume that there are no indeterminations:
$x > 0 \land y > 0 \land z > 0 \land a \neq b \neq c$
And we have to prove that $k = 1$ in:
$x^{b+c−a}⋅y^{c+a−b}⋅z^{a+b−c} = k$
Using the asker equations from cross multiplication:
$x^{b+c−a}⋅x^{a-c}⋅y^{a}⋅z^{a}⋅x^{a-b} = x^a⋅y^a⋅z^a = (xyz)^a = k$
From this method, we can also obtain:
$(xyz)^b = k$
and
$(xyz)^c = k$
But then:
$k^{\frac{1}a} = k^{\frac{1}b} = k^{\frac{1}c} \implies k = 1$
What's more:
$xyz = 1$
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