# 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 $$\log 2$$.

• Or written as $Entropy=log_2 (Phrases)$ this – kelalaka Nov 4 '18 at 20:30
• 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. – Bakuriu Nov 4 '18 at 21:58

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.