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I was wondering if $\log_1(1)$ could have more than one value, due to the fact that in the equation $1^x = 1$, $x$ can be $0$ and $1$.

Thanks in advance.

Blue
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Gabriel Accetta
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    $x$ can be any number – Anixx Jun 22 '22 at 00:00
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    I don't think it makes much sense to use the base of 1 in this context. – PC1 Jun 22 '22 at 00:02
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    "log(1) to the base of 1" There is no such thing. The logarithm function is defined for a positive base $b \ne 1$, so it makes no sense to speak of a "base of 1". – dxiv Jun 22 '22 at 00:02
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    One of the frequently forgotten rules of logarithms $\frac {\log x}{\log a} = \log_a x.$ If $a = 1$ we have a dividing by zero problem. Logs base 1 are not allowed. – user317176 Jun 22 '22 at 00:04
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    Notice also that $1^x = y$ has no solutions when $y \neq 1$. So if we try the usual definition of $\log_1 x$, we get a "function" which is only defined at one point, and at that one point results in all values - not at all useful. – aschepler Jun 22 '22 at 00:04

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The comments imply that logarithms to the base $1$ are meaningless.

We can, however, consider logarithms to the base $-1$, to a limited degree. In the real domain such logarithms can take only the arguments $+1$ and $-1$, making this function actually a duscrete logarithm. Then from the familiar powers of $-1$ we render

$\log_{-1}(+1)=\text{even}\equiv0\bmod 2$

$\log_{-1}(-1)=\text{odd}\equiv1\bmod 2$

Then the familiar product relations of $-1$ and $+1$ match up with the priduct-sum relations for logaritims. Thus

$\log_{-1}((+1)×(-1))=\text{even+odd}=\text{odd}=\log_{-1}(-1).$

We may implicitly identify an operation ○, distributive over multiplication (and also distributive under multiolicarion), with domain $\{-1,+1\}$ implicitly:

$\log_{-1}(a○b)=\log_{-1}(a)×\log_{-1}(b)$

$(+1)○(+1)=(+1)○(-1)=(-1)○(+1)=(+1)$

$(-1)○(-1)=(-1).$

As an example, we may identify each odd prime number with residue $+1$ or $-1\bmod 4$, then after this identification the quadratic reciprocity law may be expressed compactly as

$(q|p)=[(p\bmod 4)○(q\bmod 4)](p|q).$

Oscar Lanzi
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  • What you are discussing is rather discrete logarithm in a group with 2 elements (written multiplicatively). – emacs drives me nuts Jun 22 '22 at 11:18
  • I put that in. A little background: I was looking for an operation that distributes over multiplication. My ○ operation can be defined with positive bases, but I find the base $-1$ case more fascinating. – Oscar Lanzi Jun 22 '22 at 11:28