A triangle in the coordinate plane has vertices $A(\log_31,\log_33)$, $B(\log_32,\log_39)$, and $C(\log_35,\log_3{27})$. What is the area of $\triangle ABC$?

1. The vertices are $A(0,1)$, $B(\log_32,\,2)$, $C(\log_35,\,3)$.
2. By the shoelace formula (15),
   \begin{align*}
   [ABC]&=\frac12\left|0\cdot(2-3)+\log_32\cdot(3-1)+\log_35\cdot(1-2)\right|\\
   &=\frac12\left|2\log_32-\log_35\right|=\frac12\log_3\frac54 .
   \end{align*}
3. Hence $[ABC]=\boxed{\log_3\dfrac{\sqrt5}{2}}$.