algebra precalculus - How to determine answers to problems dealing with undefined values?

So we know that undefined values exist, like $\ln0$. However, if you have $e^{\ln0}$, it brings a problem where you can get $0$, if you understand that $e$ and $\ln{x}$ undo each other, or you will get undefined, because $\ln0$ is undefined and e to an undefined value is undefined. So which is correct?

Another problem I have is with exponents. We know for example that if $a-b=c$, $\dfrac{n^a}{n^b}=n^c$ which works for example, if a is $3$, b is $2$, and n is $2$, so $\dfrac{2^3}{2^2}=2$. But if $n$ is $0$, $0^{a-b}$ is the same as $\dfrac{0^a}{0^b}$, which both values are $0$ and the answer is undefined (or if you want to say $\dfrac{0}{0}$ is $1$, $1$ still isn't $0$), which would show that $0^c$ (where $c$ is any number) is undefined, since all numbers can be represented as a difference of two numbers.

In short:
My question is when do you simplify problems with undefined values (e.g $\dfrac{0^2}{0^1}$ is not $\dfrac{0}{0}$ but instead just $0^1$) and when is leaving it undefined correct? (e.g $e^{\ln0}$ being undefined, not $0$)

Answer

By the traditional meaning of equality, $e^{\ln0}$ is undefined. You cannot 'cancel' $e^x$ and $\ln{x}$ because $\ln{0}$ itself is undefined. It's analogous to rational functions like this:

$$\dfrac{(x-4)(x^2+x+3)}{(x-4)}$$

In the above function, you can cancel the term $(x-4)$ on both the top and the bottom, but that doesn't change the fact the original function has a hole at $x=4$ because you're dividing by zero. Same concept here: you can 'cancel' $e^x$ and $\ln{x}$, but that doesn't change the fact that it's undefined.

As you begin to venture into calculus and observe function behavior as $x$ approaches some constant $c$, you can use limits to state that:

$$\lim_{x\to0}\ln{x}=-\infty \Rightarrow \lim_{x\to-\infty}e^x=0$$

So as $x$ approaches $0$ in $e^{\ln{x}}$, the expression tends to $0$, but isn't equal in the traditional sense.

As for your concern about the power subtraction rule, it's simply restricted to nonzero values:

$$\dfrac{x^a}{x^b} = x^{a-b} \;\;\; x\neq0$$

So it's invalid to say:

$$\dfrac{0^2}{0^1} = 0^1 = 0$$