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, though, it would send 1 to 0, because ${\log }_b\left(1\right)=0$ for every $b$. It would also send 1 to 1, because ${\log }_b\left(b\right)=1$ for every $b$. It would also send 1 to every number at once, because $\log (1\cdot 1)$ $=$ $\log \left(1\right)+\log \left(1\right)$ and so on; this illustrates some of the difficulties inherent with logarithms base 1. And that's just when the input is 1. When the input is $x>1,$ the output would have to be $\infty ,$ and when the input is $x<1,$ the output would have to be $-\infty .$ 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 $R^{+}$ to $R\cup \{\infty ,-\infty \}.$ On the input $1$ it outputs $R$. On every input $x>1$ it outputs $\left\{\infty \right\}$. 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 $\in$, $1^{\left\{\infty \right\}}$ as the interval $(1,\infty )$, and $1^{\{-\infty \}}$ as the interval $(0,1)$. For example, $0\in {\log }_1\left(1\right)$, $1\in {\log }_1\left(1\right)$, and ${\log }_1\left(1\right)+{\log }_1\left(1\right)\in {\log }_1(1\cdot 1)$. $7\in lo{g}_1\left(1^7\right)$, and $15\in 1^{{\log }_1\left(15\right)}$. 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.