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text: '[summary: There is no [-3nd] 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.\n\nBut _if there was,_ it would be a [-multifunction] with values in the [extended_reals extended real numbers]. This is actually a perfectly valid way to define $\\log_1,$ though doing so is not necessarily a good idea.]\n\nThere is no [-3nd] 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.\n\nBut if you _really_ want a logarithm base $1$, you can define $\\log_1$ to be a multifunction [3js from] [positive_real_numebrs $\\mathbb R^+$] [3lg 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 [4bz properties of the logarithm], if you interpret $=$ as $\\in$, $1^{\\{\\infty\\}}$ as the [interval_notation 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 [complex_log final form] of the logarithm was already a multifunction, so whatever. See also [log_is_a_multifunction].\n\nWhile you're thinking about weird logarithms, see also [4c8].',
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