Logarithm base 1

https://arbital.com/p/log_base_1

by Nate Soares Jun 6 2016 updated Sep 15 2016

There is no log 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.