Key Entropy meaning Help Can anyone help me understand key entropy the mathematics behind it and the uses for it?
I understand most of it but I just want to understand what does $\log 2$ do like if I do. $\log 128 / \log 2 = 7$?

The Professor has really confused me here on this one on what the $\log 2$ mean? formula;
$Entropy = \log(Phrases)/\log 2$

The professor gave this formula, too: How many bits can represent $X$ phases? Just take the $\log X$, and divide by $\log 2$.
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.