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Can anyone help me understand key entropy the mathematics behind it and the uses for it? Thank you.

P.S please don't ignore this I need this for my preparation for my Cryptography exam in two days. I just want someone to explain this to me since I have ASD. I understand most of it but I just want to understand what does $log 2$ do like if I do. $log 128 / log 2 = 7$?

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  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: $textEntropy = log(textPhrases)/log 2$

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 $xleq 2^n$. So we first compute a logarithm, to an arbitrary base, on both sides of the inequality, which yields $log xleq log(2^n)=ncdot log 2Leftrightarrow log x/log 2leq n$, thus $log x/log 2$ bits are sufficient.