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.

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