If $x = (\log_82)^{(\log_28)}$, then $\log_3x$ equals:
$\text{(A)} \ - 3 \qquad \text{(B)} \ - \frac13 \qquad \text{(C)} \ \frac13 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 9$
By definition, $\log_8 2 = \frac 13$ and $\log_2 8 = 3$, so $\log_3 \left(\frac{1}{3}\right)^3 = \log_3 3^{-3} = -3 \Rightarrow \mathrm{(A)}$.