Question

what's the domain of y=log(x) (assuming that log is base 0)

Answer 1

Base 0?

Answer 2

\[y = \log_{0}(x)\]

Answer 3

should the domain be x = {0}

Answer 4

Ask the question: "To what power do I have to raise zero to get x?"

Answer 5

That doesn't find the domain.

Answer 6

0 ^ n = 0

Answer 7

Yes.
\[\log_0(x):0\to\mathbb{C}\backslash\{0\}\]

Answer 8

You can raise 0 to the anything, but not 0, or negative.

Answer 9

why can't you raise 0 to 0?

Answer 10

Sh*ite: \(\mathbb{C^+}\)

Answer 11

@LolWolf what does log0(x):0→C∖{0}mean?

Answer 12

\[0^0 \rightarrow \text{doesn't exist}, 0^{-n} \rightarrow \text{ divide by 0}\]

Answer 13

It means that, \(\log_0(x)\) will 'map' (has a domain of... roughly speaking) 0 to all of the positive complex numbers, excluding zero.

Answer 14

Are we doing complex exponents here?

Answer 15

nop

Answer 16

Well, the reals are part of the complex, so you can just say the same thing, in this case at least, for the reals (without including the complex numbers)

Answer 17

Then any real number other than 0 or negatives.

Answer 18

@LolWolf , why DOES maths break at 0^0?

Answer 19

math is so complicated :S

Answer 20

Because we can show \(n^0\) to mean
\[
\frac{n}{n} \text{ so, it'd be } \frac{0}{0}\text{which is undefined.}
\]

Answer 21

oh i see

Answer 22

Because if it does, surely because of:\[a^n=1a^n=a^0a^n=a^{n+0}\], so if a=0, that means that you can't raise 0 to the anythingth power

Answer 23

I know I'm wrong, but which of my equalities is the fallacious one?

Answer 24

@henpen :
\[
0^0=1
\]

Answer 25

Oh, yes- I was writing as you were explaining the previous definition

Answer 26

so to conclude, if i were to give the domain of

\[y = \log_{a}(x)\]

it would be a > 0, and x = all real numbers

Answer 27

sorry, x > 0

Answer 28

It'd be:
\[
a>0, a\neq1\\
x\in\mathbb{R}^+
\]

Answer 29

Yeah, @artofspeed , better, but make sure \(a \neq 1\)