We are given the following:
$f(n)=\prod_{d|n}g(d)$
and asked to show:
$g(n)=\prod_{d|n} f(d)^{\mu(\frac{n}{d})}$
The hint given says to use logarithms
Here's what I tried doing:
$log(f(n))=\prod_{d|n}g(d) \\ log(f(n))=log \sum_{d|n} g(d) \\ k(n)=log \sum_{d|n} g(d) \\ e^k(n)=\sum_{d|n} g(d) \\ t(n)=\sum_{d|n} g(d)$
Thus $\{t(n), g(n)\}$ are a Mobius pair. Now,
$g(n)=\sum_{d|n} \mu(d)t(\frac{n}{d}) \\ =\mu(d)e^k(\frac{n}{d}) \\ =\sum_{d|n} \mu(d)e^{log(f(\frac{n}{d}))} \\ =\sum_{d|n} \mu(d)f(\frac{n}{d})$
Not sure where to go from here, or whether this is event the correct approach to begin with.