Suppose you replace both QFTs in Shor's discrete logarithm algorithm with simpler QFTs with small prime base w. Does this algorithm extract the discrete logarithm modulo w? It seems it does, provided you guarantee that the full discrete logarithm is not too large, and that the Hadamards in the second register only generate a smaller range of values, $max(b)$ such that $max(b)max(\alpha) < p-1$, where $\alpha$ is the full discrete logarithm. So for instance, if $max(b)$ = $max(\alpha)$ = $\frac{2^{\lfloor \log p \rfloor}}{64}$, then the modified Shor's algorithm will output $\alpha \mod w$. The modified Shor circuit will end in state:
$$ \sum_{c=0}^{2^{\lceil \log p \rceil}} \sum_{d=0}^{max(b)} \sum_{a=0}^{2^{\lceil \log p \rceil}} \sum_{b=0}^{max(b)} \exp(\frac{2 \pi i}{w}(ac+bd))|c,d,g^ax^{-b} \mod p\rangle $$
The restricted range of $b$ works for the following reason. If we write $y \equiv g^k$ (the power of $x$ can be written as a power of $g$), then $a-rb \equiv k \mod (p-1)$ and
$$ a = rb + k - (p-1)\lfloor \frac{br+k}{p-1} \rfloor $$
$a$ should have the entire range of $2^{\lceil \log p \rceil}$ and $b$ should be restricted to $max(b)$. This isn't a problem because any $a$ will have a $k$ which ranges from $0$ to $p-1$, so $a$ and $b$ will always have solutions. $r$ in the substitution should be chosen to be in the range $[0,max(r)]$ in order to avoid errors with taking the second modulus.
Following Shor, the amplitude is $$ \frac{1}{w\sqrt{ max(b)max(a)}} \sum_{b=0}^{max(b)} \exp\big( \frac{2\pi i}{w}(brc+kc+bd-c(p-1)\lfloor \frac{br+k}{p-1} \rfloor)\big) $$
Factor out a factor of $\exp(2\pi i \frac{kc}{w})$ that doesn't affect the probability and get
$$ \frac{1}{w\sqrt{ max(b)max(a)}} \sum_{b=0}^{max(b)} \exp(\frac{2\pi i}{w}bT)\exp(\frac{2\pi i}{w}V) $$ where $T = rc + d - \frac{r}{p-1}\{c(p-1)\}_w$ and
$V = \big( \frac{br}{p-1} - \lfloor \frac{br+k}{p-1} \rfloor\big) \{c(p-1)\}_w$
$V$ is automatically small as in the normal algorithm. For $T$, those $c,d$ such that $rc+d=0\mod w$ encode the period modulo $w$. $\frac{\{c(p-1)\}_w}{w} < 1$, so if $\frac{max(b)r}{p-1} << 1$ then the exponentials for all terms will be close to 1 for pairs $c,d$ such that $rc+d=0\mod w$, so that the $max(b)+1$ sums will all be constructive. If $rc+d \neq 0 \mod w$, then $\exp(\frac{2\pi i}{w}bT)$ will contain terms at least as "cycled through" as $\exp(\frac{2\pi i}{w}max(b))$, and destructive interference will guarantee their sum is nearly 0.
From the Chinese Remainder Theorem, multiple runs with different small primes can be used to reconstruct the entire discrete logarithm. Note that the small prime circuit has the same asymptotic complexity as the full discrete logarithm circuit and would require O(n) runs to construct the entire logarithm, so it would be much slower in practice. All notation is from Peter Shor's original paper, except for $\alpha$.