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.

We have
$$\dfrac{\log(x)}{b-c} = \dfrac{\log(y)}{c-a} = \dfrac{\log(z)}{a-b} = t$$
This gives us
$$x=e^{t(b-c)}, y = e^{t(c-a)} \text{ and }z = e^{t(a-b)}$$
Hence,
\begin{align}
x^{b+c-a}\cdot y^{c+a-b} \cdot z^{a+b-c} & = e^{t\left((b-c)(b+c-a) + (c-a)(c+a-b) + (a-b)(a+b-c)\right)}\\
& = e^{t(b^2-c^2-ab+ac + c^2 - a^2 -bc + ba + a^2 - b^2 - ac + bc)} = e^0 = 1
\end{align}
