Product form of the Möbius inversion formula: If $f(n)>0$ for all $n$ and if $a(n)$ is real, $a(1)\neq 0$, prove that $$g(n)=\prod_{d|n}f(d)^{a(n/d)}\iff f(n)=\prod_{d|n} g(d)^{b(n/d)}$$ where $b=a^{-1}$ is the Dirichlet inverse of $a$.
So, of course, we have \begin{align*} &b(1)=\frac 1{a(1)}\\ &b(n)=\frac {-1}{a(1)}\sum_{d|n}a\left(\frac nd\right)b(d) \end{align*}
I tried taking logarithms which gave $$\log g(n)=\sum_{d|n}a(n/d)\log f(d)$$ from which I can't go any further.
On googling Product form of the Möbius inversion formula, Wikipedia gives an alternate statement $$f(n) = \prod_{d|n}g(d)\iff g(n) = \prod_{d|n} f(d)^{\mu(n/d)}?$$ which can be pretty easily proved by taking logarithms.
But, how does this relate with the original question? Am I missing something trivial?
Wikipedia's form is a specific case. Defining the unit function $u(n) = 1$, and defining the product Dirichlet multiplication by
$$\newcommand{\p}{\;\square\;} (f \p g)(n) := \prod_{d \mid n} f(d)^{g(n/d)}$$
then Wikipedia is claiming that $f = g \p u \iff g = f \p u^{-1}$, where $u^{-1}$ is taken to the usual Dirichlet inverse. One can prove that $\mu \equiv u^{-1}$.
The statement you wish to prove is the more general
$$f = g \p a \iff g = f \p a^{-1}$$
So, taking logs as you have done, we get to
$$f = g \p a \implies \log(f) = \log(g) \ast a$$
for $\ast$ the usual Dirichlet multiplication. $a$ is invertible as $a(1) \ne 0$, so we can multiply by its inverse on both sides and get
$$\log(f) \ast a^{-1} = \log(g)$$
We then convert back to something more expressive:
$$\sum_{d \mid n} \log(f(d)) a(n/d) = \sum_{d \mid n} \log \left( f(d)^{a(n/d)} \right) = \log(g)$$
Now we exponentiate and the result follows.