For a data project, I calculated a weighted average of $r$ based on the inverse weighting of $d$, a vector where all $d_i > 1$. I am interested in figuring out if it is possible to transform the resulting weighted average into the weighted average I would have gotten using a different weighting scheme, but without going back and using the full vectors of $r$ and $d$. I am not yet sure if this is possible but I have set up the problem below. I am specifically interested in backing out the weighted average I would have gotten if I took the inverse of the log of $d$ or the inverse of $d^m$.
Let
$$a = \frac{1}{\sum^N_{i=1}\frac{1}{d_i}}\sum^N_{i=1}r_i\frac{1}{d_i}$$ $$b = \frac{1}{\sum^N_{i=1}\frac{1}{\log(d_i)^m}}\sum^N_{i=1}r_i\frac{1}{\log(d_i)^m}$$ $$c = \frac{1}{\sum^N_{i=1}\frac{1}{d_i^m}}\sum^N_{i=1}r_i\frac{1}{d_i^m}$$
Find $f(\cdot)$ and $g(\cdot)$ such that,
$$ b = f(a) \ \mbox{and} \ c = g(a)$$
where $f(\cdot)$ and $g(\cdot)$ are functions that make their respective expressions true. Assume that $a$, $N$, $\overline{\frac{1}{d}}$, $\overline{r}$, and $\overline{\frac{r}{d}}$ are observable but that the individual $d_i$ and $r_i$ are unobservable. Do $f(\cdot)$ and $g(\cdot)$ exist with inputs of just $a$, $N$, $\overline{\frac{1}{d}}$, $\overline{r}$, and $\overline{\frac{r}{d}}$, for any $m\in[0,\infty)$?