So, you can say this is a random thought that popped up in my mind.
So, basically, we know that $0^n = 0 \forall n \in (0, \infty)$
And we can easily say that for $n \in {0, \infty}$, the expression is not defined by using logarithm:
$y = 0^n$
$\Rightarrow \ln{y} = \ln{0^n}$
$\Rightarrow \ln{y} = n\ln{0}$
Now, you saw it coming, for $n \rightarrow 0$, the expression becomes $$\lim\limits_{n\to0}\ln{y} = n\times\ln{0} = 0 \times \infty$$
That already is undefined, for $n \to \infty$, this becomes $$\lim\limits_{n\to\infty}\ln{y} = n\times\ln{0} = \infty\times\infty$$
Once again, both expressions are undefines and that perfectly makes sense since we already know $0^0$ and $0^\infty$ are undefines, or be precise, they're the indeterminate forms.
Again, we can say that $0^3 = 0\times0\times0 = 0$, however putting $n = 3$ in the equation above gives a shocking result: $$\ln{y} = 3\ln{0}$$ Now, again, $\ln{0}$ is undefined, right? So that makes $\ln y = undefined$.
Also, since we're talking about $0^3$, here, $0$ is not with any limit.
Now, I know that if we solve it further, we get,
$\ln{y} = 3\ln{0}$
$\to \frac{1}{3}\ln{y} = \ln{0}$
$\to \ln{y^\frac{1}{3}}$ = \ln{0}
$\to y^\frac{1}{3} = 0$
$\to y = 0$
So, okay, we started off with $y = 0^3$ and ended up at $y^\frac{1}{3} = 0$, that gives $0 = 0^3$, but that particular statement I mentioned above makes no sense. $$\ln{y} = n\ln{0}, \forall n \in (0, \infty)$$
This also holds for any value of n outside the domain of $\ln{y}$ function, for examples, negative numbers. Now, it kinda makes sense because for these values, there are only a limited conditions where they give a valid output, like for negative numbers, $\ln{y}$ makes no sense, but only if the base itself is a negative value, it makes sense. But that also exploits the property of log. For example, for some $\alpha, \beta \in R^+$, we have: $$\log_{-\beta}{-\alpha} = (-1)(-1)\log_{\beta}{\alpha} = \log_{\beta}{\alpha}$$
Which, again is a defined value since both \alpha and \beta are positive values.
This just highlights a weakness or you can say a 'blind spot' of the logarithm function. Where we get an undefined expression for a defined value. So, just out of curiosity, can someone explain me the cause of such undefined intermediates for defines values?
Now, the reason for this question is because such intermediate undefines values are not limited to only this case, but there are many cases where we get such values but they turn out to be defined values at last by simplification and using the vast mathematical properties, functions and operations. So I wanted to understand the significance of such steps by core.