# Logarithm base 1

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 $$\mathbb R^+$$ 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 $$(1, \infty)$$, 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 final form of the logarithm was already a multifunction, so whatever. See also log is a multifunction.