The Professor has really confused me here on this one on what the
$\log 2$ mean? formula: $\text{Entropy} = \log(\text{Phrases})/\log 2$
If I understand the problem correctly, you are asking what the $\log 2$ is doing there in the denominator. This is essentially to ensure that the base of the logarithm (whether it's $10$ or $e$ or $2$) doesn't matter and you always get the computation result as a base-2 logarithm.
As a quick reminder: The base in a logarithm is the $b$ for which you are looking to find the $x$ such that $b^x=a$ for $x=\log a$.
How many bits [are needed to] represent $x$ [phrases]?
So you have $x$ values and you want to know how many bits you need to have in order to identify all of them. Well, first note that 1 bit can address 2 values, 2 bits can address 4 values, 3 bits can address 8 values, etc. $n$ bits can address $2^n$ values. Thus we are looking for the smallest value $n$ such that $x\leq 2^n$. So we first compute a logarithm, to an arbitrary base, on both sides of the inequality, which yields $\log x\leq \log(2^n)=n\cdot \log 2\Leftrightarrow \log x/\log 2\leq n$, thus $\log x/\log 2$ bits are sufficient.