# 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.