Given $\frac{\log{a}}{p}=\frac{\log{b}}{q}=\frac{\log{c}}{r}=\log{x}$, all logarithms to the same base and $x \not= 1$.  If $\frac{b^2}{ac}=x^y$, then $y$ is:
$\text{(A)}\ \frac{q^2}{p+r}\qquad\text{(B)}\ \frac{p+r}{2q}\qquad\text{(C)}\ 2q-p-r\qquad\text{(D)}\ 2q-pr\qquad\text{(E)}\ q^2-pr$
We are given:
\[\frac{b^2}{ac} = x^y\]
Taking the logarithm on both sides:
\[\log{\left(\frac{b^2}{ac}\right)} = \log{x^y}\]
Using the properties of logarithms:
\[2\log{b} - \log{a} - \log{c} = y \log{x}\]
Substituting the values given in the problem statement:
\[2q \log{x} - p \log{x} - r \log{x} = y \log{x}\]
Since $x \neq 1$, dividing each side by $\log{x}$ we get:
\[y = \boxed{\textbf{(C) } 2q - p - r}\]
~ proloto