# 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.

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