A function $g(n)$ is defined for positive integers $n$ by the rule $\displaystyle\sum_{d|n}g(d) = \log n$. Prove that $g(n)=\log p$ if $n=p^e$ where $p$ is a prime and $e\in \mathbb{Z}^+$, and $g(n)=0$ otherwise.
I have no idea how to start this question at all does it include mathematical induction?
On
We have by Mobius inversion that
$$g(n) = \sum_{d|n} \mu(d) \log(n/d) = \log n \times \sum_{d|n} \mu(d) - \sum_{d|n} \mu(d) \log d \\ = - \sum_{d|n} \mu(d) \log d.$$
Now collect the contributions from $\log p$ where $p\in P$ with $P$ the set of primes that divide $n$. We obtain
$$-\sum_{p|n} \log p \sum_{Q\subseteq P\setminus \{p\}} (-1)^{|Q|+1} = \sum_{p|n} \log p \sum_{Q\subseteq P\setminus \{p\}} (-1)^{|Q|} \\ = \sum_{p|n} \log p \sum_{q=0}^{|P\setminus\{p\}|} {|P\setminus\{p\}| \choose q} (-1)^q.$$
Observe that when $|P\setminus\{p\}| \ge 1$ i.e. $n$ has more than one prime factor we get
$$\sum_{p|n} \log p \times (1-1)^{|P\setminus\{p\}|} = 0,$$
so $g(n)$ is zero in this case. This leaves $n$ the power of a single prime $p$ and we find
$$\log p \times (-1)^{|\emptyset|} = \log p$$
as claimed.
If you don't know where to begin, then you begin with small steps. Start with $g(1)$. Then tackle $g(2), g(3), g(5)$ (note that the definition says that $g(1) + g(2) = \log 2$, and so on), and from there you should be able to see quite easily that $g(p) = \log p$ for prime $p$.
Now for prime powers. Calculate $g(4)$ from $g(1) + g(2) + g(4) = \log 4$. Then $g(8)$ similarily. Here you might have to use well-known facts about the logarithm to proceed. Now you can probably prove that $g(2^e) = \log 2$, and it shouldn't be too hard to show that the same goes for all the other primes.
Finally, numbers which are not powers of primes. Start with $g(6), g(10)$ and $g(15)$ to get a feel for it. Maybe $g(12), g(18)$ and $g(30)$. Then go for the full result from there. Doing some form of (strong) induction might prove advantageous here.