In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$, and the altitude to these bases has length $\log 16$. The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$, where $p$ and $q$ are positive integers. Find $p + q$.
Call the trapezoid $ABCD$ with $AB$ as the smaller base and $CD$ as the longer. Let the point where an altitude intersects the larger base be $E$, where $E$ is closer to $D$. 
Subtract the two bases and divide to find that $ED$ is $\log 8$. The altitude can be expressed as $\frac{4}{3} \log 8$. Therefore, the two legs are $\frac{5}{3} \log 8$, or $\log 32$.
The perimeter is thus $\log 32 + \log 32 + \log 192 + \log 3$ which is $\log 2^{16} 3^2$. So $p + q = \boxed{018}$