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 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 $R^{+}$ to $R \cup {\infty, - \infty} .$ On the input $1$ it outputs $R$ . 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 ${log}_{1} (1) + {log}_{1} (1) \in {log}_{1} (1 \cdot 1)$ . $7 \in l o g_{1} (1^{7})$ , 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.