Consider the finite fields GF$(168485857)$ which has $168485856$ elements not counting the zero element. $g=5$ is a primitive root.
Let $C$ be the cyclic group which contains all elements $e$ that are $7392$-nd power residues$\pmod {168485857}$.
Let $G$ be the cyclic group which contains all elements $k$ that are $22793$-rd power residues$\pmod {168485857}$.
Find elements $e$ and $k$ from $C$ and $G$ respectively such that:
$ek=5$ in GF$(168485857)$
This is the same as finding an integers ${E, K, e, k} \pmod {168485857}$ such that $E^{7392}=e$, $K^{22793}=k$, and $ek=5 \pmod {168485857}$.
This is a discrete logarithm problem with two bases. Anyone know where to start?
This is most emphatically NOT a discrete logarithm problem. Bezout's identity suffices.
Let $p=168485857$. Then $p-1=(2^5\cdot3\cdot7\cdot11)\cdot(23\cdot991)$, where the factor inside the first parens is $u=7392$ and the one in the latter is $v=22793$. So $uv=p-1$. Furthermore, $5^u$ is a generator of the subgroup $C$, and $5^v$ is a generator of the subgroup $G$.
The first thing we need to do is to run the extended Euclidean algorithm to get $\gcd(u,v)=1$ in the Bezout identity form: $$ 1=9494u-3079v. $$ Therefore, in the field $GF(p)$, we have $$ 5=5^1=5^{9494u-3079v}=(5^u)^{9494}\cdot(5^v)^{-3079} $$ Here the first factor $(5^u)^{9494}$ is an element of $C$ and the second factor $(5^v)^{-3079}=(5^v)^{4313}$ is an element of the subgroup $G$. I used the fact that $5^v$ is of order $u$, so as $-3079\equiv4313\pmod{u}$ we can replace $-3079$ with a positive exponent.
Mathematica then helps me (I could have used square-and-multiply instead) with $$ \begin{aligned} e=(5^u)^{9494}&\equiv 101471627\pmod p,\\ k=(5^v)^{4313}&\equiv 99287708\pmod p. \end{aligned} $$