Through the use of theorems on logarithms 
\[\log{\frac{a}{b}} + \log{\frac{b}{c}} + \log{\frac{c}{d}} - \log{\frac{ay}{dx}}\]
can be reduced to: 
$\textbf{(A)}\ \log{\frac{y}{x}}\qquad  \textbf{(B)}\ \log{\frac{x}{y}}\qquad  \textbf{(C)}\ 1\qquad \\ \textbf{(D)}\ 140x-24x^2+x^3\qquad \textbf{(E)}\ \text{none of these}$
Using the properties $\log(x)+\log(y)=\log(xy)$ and $\log(x)-\log(y)=\log(x/y)$, we have
\begin{align*} \log\frac ab+\log\frac bc+\log\frac cd-\log\frac{ay}{dx}&=\log\left(\frac ab\cdot\frac bc\cdot\frac cd\right)-\log \frac {ay}{dx} \\ &=\log \frac ad-\log\frac{ay}{dx} \\ &=\log\left(\frac{\frac ad}{\frac{ay}{dx}}\right) \\ &=\log \frac xy, \end{align*}
so the answer is $\boxed{\textbf{(B)} \log\frac xy}.$