I am asking about the dos and don'ts of piecewise expressions. First of all, there's the question of completeness versus clarity. Being as complete as possible can lead to long piecewise expressions that may only add a little info; that info may even be a little obvious as well, meaning one could say it's unnecessary. In the piecewise expression below, the first three cases detail what variables lead to undefined answers when plugged into the log function. Perhaps a better way to show this would be to simply state this beforehand, and then omit those disallowed variables from the piecewise function?
Here's the options:
Applies to all of the identities: $a^n \in \Bbb R \lor a^n = \text{UNDEF}$ -- it's either real, or undefined; no complex or imaginary values.
$\tag 1$ $$\log_b a^n = \begin{cases} \text{UNDEF} & \qquad b \in (- \infty, 1], \ \ a,n \in (- \infty, \infty) \\[2ex] \text{UNDEF} & \qquad b \in (1, \infty) , \ \ a^n \in (-\infty, 0], \ \ n \neq 0 \\[2ex] \text{UNDEF} \lor 0 & \qquad b \in (1, \infty), \ \ a=n=0\\[2ex] 0 & \qquad b \in (1, \infty), \ \ a^n = 1 \\[2ex] 1 & \qquad b=a^n \in (1, \infty) \\[2ex] n & \qquad b=a \in (1, \infty), \ \ n \in (-\infty, \infty) \\[2ex] x \in (-\infty, \infty) & \qquad b,a \in (1, \infty), \ \ n \in (-\infty, \infty) \end{cases}$$
Or:
The base and argument of a log function cannot be less than zero, and the base cannot be one. $\tag 2$ $$\log_b a^n = \begin{cases} \text{UNDEF} \lor 0 & \qquad b \in (1, \infty), \ \ a=n=0\\[2ex] 0 & \qquad b \in (1, \infty), \ \ a^n = 1 \\[2ex] 1 & \qquad b=a^n \in (1, \infty) \\[2ex] n & \qquad b=a \in (1, \infty), \ \ n \in (-\infty, \infty) \\[2ex] x \in (-\infty, \infty) & \qquad b,a \in (1, \infty), \ \ n \in (-\infty, \infty) \end{cases}$$
Now, the top case here could be considered obvious and redundant. It is basically saying that if the argument ($a^n$, not $a$) is equal to one, then the answer must be equal to $0$, which is stated in the case below it. Thus, it is really just repeating the info of the case below it, only it is less general. However, since there's two possible answers to the value of $0^0$, it is distinguished from the case below. So, is this kind of case even necessary? Perhaps the best of these three would be:
$\tag 3$ $$\log_b a^n = \begin{cases} 0 & \qquad b \in (1, \infty), \ \ a^n = 1 \\[2ex] 1 & \qquad b=a^n \in (1, \infty) \\[2ex] n & \qquad b=a \in (1, \infty), \ \ n \in (-\infty, \infty) \\[2ex] x \in (-\infty, \infty) & \qquad b,a \in (1, \infty), \ \ n \in (-\infty, \infty) \end{cases}$$
Lastly, the question of what variables to show the value of. On the LHS, there is four variables: $b, a, n, a^n$. The value of $a^n$, or what set it belongs to, depends on the values/sets of $a$ and $n$. When one defines $a$ and $n$, it can create a kind of imagined piecewise function for the possible values of $a^n$, and vice versa. Is this okay, or should all the variables be present in every case? Here's what I mean:
$\tag 4$ $$\log_b a^n = \begin{cases} \text{UNDEF} & \qquad b \in (- \infty, 1], \ \ a,n, \in (- \infty, \infty), \ \ a^n \in (-\infty, \infty) \lor a^n = \text{UNDEF} \\[2ex] \text{UNDEF} & \qquad b \in (1, \infty) , \ \ a^n \in (-\infty, 0], \ \ a = 0 \neq n \lor (a \in (-\infty, 0) \land n \in \Bbb Z)\\[2ex] \text{UNDEF} \lor 0 & \qquad b \in (1, \infty), \ \ a=n=0, \ \ a^n = 1 \lor a^n = \text{UNDEF} \\[2ex] 0 & \qquad b \in (1, \infty), \ \ a^n = 1, \ \ n=0 \neq a \\[2ex] 1 & \qquad b=a^n \in (1, \infty), \ \ a,n \in (1, \infty) \\[2ex] n & \qquad b=a \in (1, \infty), \ \ n \in (-\infty, \infty), \ \ a^n \in (-\infty, \infty) \\[2ex] x \in (-\infty, \infty) & \qquad b,a,a^n \in (1, \infty), \ \ n \in (-\infty, \infty) \end{cases}$$
To summarize the questions:
- Which of these is best; $(1), (2), (3)$?
- Does one have to include values/sets describing all the variables? Or can one include enough info so that a piecewise function/value/set for the excluded variable(s) could have been extrapolated from the shown information?
Outside of that, is there a better way to show the possible values of $\log_b a^n$ in a piecewise function not shown in this post? Perhaps a piecewise function is just a bad format for this task to begin with?