Since logarithms are monotonically increasing, you can determine if $$a^b > c^d$$ by instead checking if $$\log(a^b) > \log(c^d)$$ which can be rewritten (using basic properties of logarithms) as $$b\log(a) > d\log(c)$$ This results in the comparison of two numbers on the order of those given in the input (that is, both numbers will be within an order of magnitude or so of $a,b,c,d$), which is definitely feasible on a computer.
You can also use logs to calculate other properties, as long as you remember to convert your solution back to the original numbers (by means of an exponential) when you're done. For example, to compute whether $${a_1}^{b_1}{a_2}^{b_2} > {c_1}^{d_1}{c_2}^{d_2}$$, you can take the log of both sides to get
\begin{align}
\log({a_1}^{b_1}{a_2}^{b_2}) &> \log({c_1}^{d_1}{c_2}^{d_2}) \\
\log({a_1}^{b_1}) + \log({a_2}^{b_2}) &> \log({c_1}^{d_1}) + log({c_2}^{d_2}) \\
b_1\log(a_1) + b_2\log(a_2) &> d_1\log(c_1) + d_2\log(c_2) \\
\end{align}
which is again computationally feasible.
Here is one way to do it that requires knowing only a primitive root $\bmod p$ to get all primitive roots $\bmod p^2$, provided that $p$ is twice a smaller prime plus $1$, so that all nonzero residues besides $\pm 1$ are either quadratic or primitive. I will demonstrate it for $p=5$ letting you apply it for $p=11$.
Start with $2$ as a primitive root $\bmod 5$. Then there will be a series of primitive roots $\equiv 5k+2 \bmod 25$. For one residue $k \bmod 5$ we will have $(5k+2)^4\equiv 1 \bmod 25$ and that value of $k$ must be rejected, but other values of $k$ will all work.
Use the Binomial Theorem to get $(5k+2)^4$, but retain only the first ad zero powers of $k$ (why?). Thus
$(5k+2)^4 \equiv (4)(5k)(2^3)+2^4 \equiv 1 \bmod 25$
$10k+16 \equiv 1, 10k \equiv 10 \bmod 25$
Divide by $10$ noting that the modulus is to be divided by $gcd(10,25)=5$. Then $k\equiv 1 \bmod 5$ meaning $7$ will not be a primitive root $\bmod 25$ but all other residues two greater than a multiple of $5$ will work. We therefore get a subset of primitive roots:
$\{2,12,17,22\}$
Now take the root we rejected, $7$. If we multiply the above primitive roots by it, we get nonprimitive roots because these are quadratic residues. But multiply by $7$ twice, that is by $7^2$, and we get another set of primitive roots because the products are non-quadratic and also miss being $\equiv -1 \bmod 5$. Since $7^2 \equiv 24 \equiv -1 \bmod 25$ this gives another subset of primitive roots
$\{3,8,13,23\}$
Multiplying by $7^2$ again cycles us back to the first subset, so the complete set of primitive roots $\bmod 25$ will be the union of the two subsets
$\{2,3,8,12,13,17,22,23\}$.
Now apply this method to$11=2×5+1$ and see what happens. Since $11$, like $5$, is not one greater or one less than a multiple of $8$, $2$ will be a primitive root $\bmod 11$. Since there are four such primitive roots $\bmod 11$ overall you will need to generate four subsets before taking their union.
Best Answer
In general, if $g$ is a primitive root mod $p$ then $g$ or $g+p$ is a primitive root mod $p^2$.
For $p=11$, we have that $g=2$ works.
It is enough to test that $2^{110/q}\not \equiv 1 \bmod 121$ for $q=2,5,11$, the prime divisors of $110=\phi(121)$.