What is the significance of using the term "discrete" in discrete logarithm? I'm trying to clear up my confusion in using the term "discrete" in discrete logarithm. I'm focusing on why the word "discrete" is used to differentiate it from a logarithm. 
Wikipedia defines a discrete logarithm as follows: 

in any group $G$, powers $b^k$ can be defined for all integers $k$,
  and the discrete logarithm $log_b a$ is an integer $k$ such that $b^k
= a$.

Is the term discrete added simply to reflect the fact that $k$ in $log_ba=k$ is confined to integers? Or is it a combination of the discrete log being an integer as well as the powers fulfilling the properties of a group $G$?
 A: From Google, the definition of discrete is:

individually separate and distinct.

In the math sense, this means only taking on a specific set of values. In modular arithmetic these are the integers. This applies to the exponents, as much as it applies to the remainders. Also though the normal log can help in some sense, it doesn't wrap around like a discrete log. About the only way in which the normal log gives us an answer to the discrete log is if we take the fractional part as a fraction defined properly in mod $\varphi(p)$ math. And even then it's not guaranteed correct. 
A: I'm not sure I understand your dichotomy, but because the generating set in question is discrete, the logarithm is called a discrete logarithm. Here discrete is used to distinguish the logarithm from a continuous logarithm, which would be a logarithm in the real numbers or a similarly continuous set. 
This need not happen in a finite group, as long as the group is discrete itself. In a general group, one composes an element with itself a set number of times to achieve a value $k$ (usually by binary exponentiation). When this process is reversed, the values that can be taken by the powers $b^k$ in the set generated by $b$ are discretized, since you can only have integer powers of $b$ in a black box group.
The reason discrete is mentioned is to separate the problem from a similar problem in a continuous case with a continuous set. I we supposed a "continuous logarithm problem" existed it could easily be solved computationally, since one can simply compute the log of both sides of the equation, which I think boils down to Newton's method. Anyways it's fair to say a computer can estimate it quite accurately without trouble.
