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.

If you want to use your equations, here is a method.
Multiplying the equations together, we obtain: $$x^{c-a}y^{a-b}z^{b-c}=y^{b-c}z^{c-a}x^{a-b}$$ which gives after reordering: $$x^{b+c-a}y^{c+a-b}z^{a+b-c}=x^a y^b z^c$$
Therefore it suffices to show that $x^a y^b z^c = 1$. 
Your first and third equations give $y = x^{\frac {c-a}{b-c}}, z = x^{\frac{a-b}{b-c}}$. This gives us: 
$$x^a y^b z^c = x^a x^{\frac {c-a}{b-c}\times b} x^{\frac{a-b}{b-c}\times c} = x^{a + \frac{bc-ba+ca-bc}{b-c}} = x^{a-a} = x^0 = 1 $$
QED.
