Prove $\Lambda (n)=-\sum_{d | n} \mu(d) \log(d)$ Prove $\Lambda (n)=-\sum_{d | n} \mu(d) \log(d)$
I'm not sure where to begin with this. $\Lambda (n)$ is the Von Mangodt function that is defined by $\Lambda (n) = \log p$ if n is a power of the prime p and $0$ otherwise.
Progress:
Start by showing $\sum_{d | n}\Lambda = \log(n)$
For $n=1$ it holds since $\Lambda(1)=0=\log 1$. For $n$ greater or equal to $2$, we have $\sum_{d | n}\Lambda = \sum_{p^m | n}\log p = \log(n)$
 A: First show that, for all $n$:
$$\log n = \sum_{d\mid n} \Lambda(d)$$
So, by Möbius inversion:
$$\begin{align}\Lambda (n) &=\sum_{d\mid n}\mu(d)\log\left(\frac n d\right)\\
&=\sum_{d\mid n}\mu(d)\left(\log n -\log d\right)\\
&=\log n\sum_{d\mid n}\mu(d)-\sum_{d\mid n}\mu(d)\log d\\
&=-\sum_{d\mid n}\mu(d)\log d
\end{align}$$
The last step because, when $n>1$, $\sum_{d\mid n}\mu(d)=0$, and when $n=1$, $\log n=0$.
A: This is the long proof, but also the most interesting. For $\Re s>1$
$$\zeta(s) = \sum_{n=1}^\infty n^{-s} = \prod_p (1+\sum_{k \ge 1}p^{-sk}) = \prod_p \frac{1}{1-p^{-s}}.$$
So that 
$$\ln \zeta(s) = -\sum_p \ln(1-p^{-s}) = \sum_{p}\sum_{k \ge 1} \frac{p^{-sk}}{k}$$
and
$$\frac{\zeta'(s)}{\zeta(s)} = \frac{d}{ds} \ln \zeta(s) =-\sum_{p}\sum_{k \ge 1} \frac{p^{-sk}}{k}\ln p^k = -\sum_{n=1}^\infty \Lambda(n) n^{-s}.$$
On the other hand
$$\frac{1}{\zeta(s)} = \prod_p(1-p^{-s}) = \sum_{n=1}^\infty \mu(n) n^{-s}, \qquad \frac{-\zeta'(s)}{\zeta(s)^2} = \frac{d}{ds}\frac{1}{\zeta(s)}= -\sum_{n=1}^\infty \mu(n) n^{-s}\ln(n)$$
and hence
$$\frac{\zeta'(s)}{\zeta(s)}=\frac{\zeta'(s)}{\zeta(s)^2} \zeta(s) = (\sum_{n=1}^\infty \mu(n) n^{-s}\ln(n))(\sum_{n=1}^\infty n^{-s}) = \sum_{n=1}^\infty n^{-s} \sum_{d | n} \mu(d) \ln(d)$$
i.e.
$$\Lambda(n) = -\sum_{d | n} \mu(d) \ln(d).$$
