Prove $\lfloor \log_2(n) \rfloor + 1 = \lceil \log_2(n+1) \rceil $ This is a question a lecturer gave me. I'm more than willing to come up with the answer. But I feel I'm missing something in logs.
I know the rules, $\log(ab) = \log(a) + \log(b)$ but that's all I have. 
What should I read, look up to come up with the answer?

Well, even after your edit, this is not an identity in general. It is valid only for integral $n$. For a counter-example of why this is not valid for a positive $n$, take $n = 1.5$. Then,
$$\lfloor \log_2(n) \rfloor + 1 = \lfloor \log_2(1.5) \rfloor + 1 = 1$$
and
$$\lceil \log_2(n + 1) \rceil = \lceil \log_2(2.5) \rceil = 2$$
However, for $n \in \mathbb N$ and $n > 0$, this identity holds, and can be proven as follows:
Let $2^k \le n < 2^{k+1}$ for some $k \ge 0$ and $k$ is an integer.
Therefore, let $n = 2^k + m$, where $k \ge 0$ and $0 \le m < 2^{k}$, and $k, m$ are positive integers.
Then, 
$$\lfloor \log_2(n) \rfloor + 1 = \lfloor \log_2(2^k + m) \rfloor + 1 = k + 1$$
Notice that $\log_2(2^k + m) = k$ when $m = 0$. Otherwise, $k < \log_2(2^k + m) < k+1$, and hence the above result.
Also, 
$$\lceil \log_2(n + 1) \rceil = \lceil \log_2(2^k + m + 1) \rceil = k + 1$$
This last equation is clear for $m < 2^k -1$, where $2^k + m + 1 < 2^{k+1}$. When $m=2^k-1$, notice that $\log_2(2^k+2^k-1+1) = \log_2(2^{k+1})$ is an integer, and therefore $\lceil k+1\rceil=k+1$.
Hence, this identity is valid only for positive integers ($n \in \mathbb N$).
