Can't understand trivial discrete logarithm problem for $Z_{11}$ I have a seemingly trivial problem with description:

Find all discrete logarithms of base 2 of all non-zero elements in
  $Z_{11}$ field.

I'm basing my learning on the notes I managed to grab from other groups and for that problem, there's a solution with only 5 $dlog$'s calculated, namely $dlog_2 1, dlog_2 2, dlog_2 10, dlog_2 7, dlog_2 3$. What I don't get is why are all the other elements of $Z_{11}$ ignored? 2 is a generator of the field so for different values of $i$ you could derive such $2^i$ that it yields numbers from $1$ to $10$. Why don't we calculate the $dlog$'s of all such numbers but only half of them? Or is there  mistake in the answer?

If $x\cdot y \equiv 1 \mod p$, then $$
  \underbrace{\textrm{dlog}_2 x}_{=u} + \underbrace{\textrm{dlog}_2 y}_{=v} \equiv 0 \mod (p-1)
$$
because if $2^u \equiv x, 2^v \equiv y$ then $x\cdot y \equiv 2^{u+v} \equiv 1$ if $u + v \mid 10$ (this follows from fermat's little theorem). Over $\mathbb{Z}_{11}$ you have $$\begin{eqnarray}
  1 &\cdot& 1 &\equiv& 1 \\
  2 &\cdot& 6 &\equiv& 1 \\
  3 &\cdot& 4 &\equiv& 1 \\
  5 &\cdot& 9 &\equiv& 1 \\
  7 &\cdot& 8 &\equiv& 1 \\
  10 &\cdot& 10 &\equiv& 1
\end{eqnarray}$$
meaning that the discrete $2$-logarithmns of $1,2,3,5,7,10$ suffice to easily express the others. Why your list is missing $\textrm{dlog}_2 5$, I don't know.
