0

$\log _{ 1 }{ 1 } =1\quad $ v $0$?

Because $1^1 = 1$ and $1^0 = 1$?

haqnatural
  • 21,578
Xiaox
  • 1

1 Answers1

9

There's no such thing as base-$1$ logarithms (because $x\mapsto 1^x$ is not injective, if you want a concrete reason), so claiming anything about it is incorrect.

hmakholm left over Monica
  • 286,031
  • I know that log_a (b) exists only (in R) if a&b>0 and a =/= 1 but im curious why it ain't true if def. clearly says that log_a(b) = c <==> a^x = b (sorry if u cannot understand me properly...) – Xiaox Jun 07 '16 at 20:25
  • 1
    @Xiaox: If you have a source that claims that to be the definition without requiring that $a\ne 1$, then that's a source that peddles bad definitions, and you should treat what it says with skepticism. – hmakholm left over Monica Jun 07 '16 at 20:26
  • The definition that I know requires a ≠ 1, but as I said before I'm curious why there isn't log_1 (1) if log_a(b) = c <==> a^x = b – Xiaox Jun 07 '16 at 20:31
  • @Xiaox: That definition will not work for $a=1$, because a definition of a function is supposed to define one and only one function value for each argument -- but since $1=1^0=1^1=1^2=1^{-42}=1^\pi=\cdots$ there are a lot of different values that your attempted definition all claim equal $\log_1(1)$. But it's implicit in the use of function notation the $\log_1(1)$ should denote only one number (well, at most one number, because it actuall denotes no number at all) -- so a "definition" that would lead it to equal everything actually fails to define anything at all. – hmakholm left over Monica Jun 07 '16 at 20:36
  • Well, thanks. PS: It may be considered as exception? – Xiaox Jun 07 '16 at 20:40