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.

The product of the three equalities you have given is:
$x^{c-a} y^{a-b} z^{b-c} = y^{b-c} z^{c-a} x^{a-b}$

Grouping similar variables together,
$(x^{c-a} x^{b-a}) (y^{a-b} y^{c-b}) (z^{b-c} z^{a-c}) = 1$
$x^{-2a+b+c} \times y^{a-2b+c} \times z^{a+b-2c} = 1$

