Vtyjkyuk

A function $g(n)$ is defined for positive integers $n$ by the rule
$displaystylesum_dg(d) = log n$.
Prove that $g(n)=log p$ if $n=p^e$ where $p$ is a prime and $ein mathbbZ^+$, and $g(n)=0$ otherwise.

I have no idea how to start this question at all does it include mathematical induction?

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.

We have by Mobius inversion that

$$g(n) = sum_d mu(d) log(n/d)
= log n times sum_d mu(d) - sum_d mu(d) log d
\ = - sum_d mu(d) log d.$$

Now collect the contributions from $log p$ where $pin P$
with $P$ the set of primes that divide $n$. We obtain

$$-sum_p log p
sum_Qsubseteq Psetminus p (-1)^Q
= sum_p log p
sum_Qsubseteq Psetminus p (-1)^
\ = sum_p log p
sum_q=0^
(-1)^q.$$

Observe that when $|Psetminusp| ge 1$ i.e. $n$ has more than
one prime factor we get

$$sum_p log p times (1-1)^ = 0,$$

so $g(n)$ is zero in this case. This leaves $n$ the power of a single
prime $p$ and we find

$$log p times (-1)^ = log p$$

as claimed.