Logarithm base 1

[summary: There is no Logarithm base 1, because no matter how many times you multiply 1 by 1, you get 1. If there were a log base 1, it would send 1 to 0 (because $\log_b(1)=0$ for every $b$), and it would also send 1 to 1 (because $\log_b(b)=1$ for every $b$), which demonstrates some of the difficulties with $\log_1.$ In fact, it would need to send 1 to every number, because $\log(1 \cdot 1) = \log(1) + \log(1)$ and so on. And it would need to send every $x > 1$ to $\infty$, and every $0 < x < 1$ to $-\infty,$ and those aren't numbers, so there's no logarithm base 1.

But if there was, it would be a [-multifunction] with values in the [extended_reals extended real numbers]. This is actually a perfectly valid way to define $\log_1,$ though doing so is not necessarily a good idea.]

There is no Logarithm base 1, because no matter how many times you multiply 1 by 1, you get 1. If there were a log base 1, it would send 1 to 0 (because $\log_b(1)=0$ for every $b$), and it would also send 1 to 1 (because $\log_b(b)=1$ for every $b$), which demonstrates some of the difficulties with $\log_1.$ In fact, it would need to send 1 to every number, because $\log(1 \cdot 1) = \log(1) + \log(1)$ and so on. And it would need to send every $x > 1$ to $\infty$, and every $0 < x < 1$ to $-\infty,$ and those aren't numbers, so there's no logarithm base 1.

But if you really want a logarithm base $1$, you can define $\log_1$ to be a multifunction from [positiverealnumebrs $\mathbb R^+$] to $\mathbb R \cup \{ \infty, -\infty \}.$ On the input $1$ it outputs $\mathbb R$. On every input $x > 1$ it outputs $\{ \infty \}$. On every input $0 < x < 1$ it outputs $\{ -\infty \}$. This multifunction can be made to satisfy all the basic properties of the logarithm, if you interpret $=$ as $\in$, $1^{\{\infty\}}$ as the [interval_notation interval] $(1, \infty)$, and $1^{\{-\infty\}}$ as the interval $(0, 1)$. For example, $0 \in \log_1(1)$, $1 \in \log_1(1)$, and $\log_1(1) + \log_1(1) \in \log_1(1 \cdot 1)$. $7 \in log_1(1^7)$, and $15 \in 1^{\log_1(15)}$. This is not necessarily the best idea ever, but hey, the [complex_log final form] of the logarithm was already a multifunction, so whatever. See also [log_is_a_multifunction].

While you're thinking about weird logarithms, see also Log base infinity.