I want to prove the natural log can't equal 0. Is this done via the limit definition or the fact the inverse of ln is e and e can't be 0?
How would this proof be started as well thanks.
I mean, this is all kind of semantical. I could call $\displaystyle \ln(0)=e$ if I wanted to. If you're asking though, given the predefined function $\displaystyle \exp:\mathbb{R}\to\mathbb{R}^+$ then the inverse function $\displaystyle \exp^{-1}$ has domain equal to the codomain of $\displaystyle \exp$, namely $\displaystyle \mathbb{R}^+$ and since $\displaystyle 0\notin\mathbb{R}^+$, we have that $\displaystyle \exp^{-1}$ isn't (at least not with keeping it as an inverse of the exponential function) defined at zero.
But that doesn't make sense. If you don't define $\displaystyle \ln$ as being the inverse of $\displaystyle \exp$ then I'd ask you 'what is it' and no matter how you've answered (assuming you've done so rigorously) you will, by necessity, have answered your own question. You can't define $\displaystyle \ln$ without specifying its domain. A function can be thought of as the ordered triple $\displaystyle \left(A,B,f\right)$ where $\displaystyle A,B$ are sets and $\displaystyle f\subseteq A\times B$ is a relation such that for each $\displaystyle a\in A$ there is precisely one $\displaystyle b\in B$ for which $\displaystyle (a,b)\in f$.
But that isn't really well-defined. Really what you may want to consider is that $\displaystyle \displaystyle \ln(x)=\int_1^x \frac{dt}{t}$ and that if you take the domain to the largest for which that equation makes sense then surely $\displaystyle 0$ isn't included. But that's really quite silly.
When I see something like this it convinces me that indefinite integration should never be taught.
That does not define the natural logarithm. It leaves it defined up to an additive constant. You need to define it as a definite integral when the non-convergence of that integral when the upper limit is zero shows that the natural log of zero does not exist.
CB
That is more exact to say that...
$\displaystyle \displaystyle \ln |t| = \left\{\begin{array}{ll}\int_{-1}^{t} \frac{d \tau}{\tau} ,\,\,t<0\\{}\\ \int_{1}^{t} \frac{d \tau}{\tau} ,\,\, t > 0\end{array}\right.$
In any case the function $\displaystyle \ln |t| $ has a singularity in $\displaystyle t=0$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Take the very basic, high-schoolish, definition of logarithm: $\displaystyle \log_ab=x\Longleftrightarrow a^x=b$ , when we already
know-define-demmand that $\displaystyle a,b>0\,,\,a\neq 1$ . From this, $\displaystyle b=0\Longrightarrow a^x=0$ for some
$\displaystyle x\in\mathbb{R}$ which, as Drexel already pointed out, is impossible by the very
definition of power of positive real numbers (although he talked about inverse function and
stuff, which isn't necessary in a first approach to this)
Tonio
In other words it makes no sense to try to proved anything about a function without saying how you have defined it! One way to show that 0 is not in the domain of ln(x) is to define ln(x) is as the inverse function of $\displaystyle f(x)= e^x$ and then show that $\displaystyle e^x= 0$ is not true for any x. How you do that would, of course, depend upon how you defined $\displaystyle e^x$.
Another way to define ln(x) is
$\displaystyle ln(x)= \int_1^x \frac{1}{t}dt$
You can then show that 0 is not in the domain of ln(x) by showing that $\displaystyle \int_1^0 \frac{1}{t}dt$ does not exist.