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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 $\log 2$.

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  • $\begingroup$ Or written as $Entropy=log_2 (Phrases)$ this $\endgroup$ – kelalaka Nov 4 '18 at 20:30
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    $\begingroup$ FYI: adding things like "I need this in two days" etc can only have a negative impact on your audience. You make yourself look as someone that uses readers just as a tool. Avoid that in future questions. $\endgroup$ – Bakuriu Nov 4 '18 at 21:58
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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.

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  • $\begingroup$ Thanks for helping, I the professor gave this formula: How many bits can represent X phases? Just take the Log(X), and divide by Log (2). $\endgroup$ – Weaponized Autism Nov 4 '18 at 20:46

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