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Are $log_1 1$ and $log_0 0$ indeterminate forms?

Whenever I ask someone about these indeterminate forms, they deny by saying either $log$ is neither defined at base $0$ nor at base $1$, or they say $log$ is a function so these must not be included in fundamental indeterminates.

But, we know division by zero is not defined, yet $0/0$ is indeterminate; and many others. And, actually, $log$ is more a binary-operator that is the inverse operation of power/exponent operator.

I think that your difficulty comes from a confusion regarding what an "indeterminate form" is. Indeterminate forms show up in analysis via naive substitution when computing limits. For example, we might naively compute
$$
lim_{xto 0} frac{x^2}{x}
= frac{lim_{xto 0} x^2}{lim_{xto 0} x}
= frac{0}{0}.
$$
Since this last expression is undefined, we might say that the limit is "indeterminate of the form $frac{0}{0}$." When this kind of naive substitution leads to an undefined expression, it is necessary to be a bit more clever in the evaluation of the limit. In this case,
$$
lim_{xto 0} frac{x^2}{x}
= lim_{xto 0} x
= 0.
$$
Techniques for working with indeterminate forms include results such as L'Hospital's rule, applying algebraic transformations, and so on.

In the case of "the logarithm base 0", $log_0(x)$ is undefined. This expression doesn't make sense. If this expression were defined, then it must be equal to some number, say $y$. Then
$$ log_0(x) = y implies x = 0^y = 0. $$
But $0^y = 0$ for any positive value of $y$. Hence the expression $log_0(x)$ is not well defined, as there is a not a unique value of $y$ which gets the job done. On the other hand, we can consider limits of expressions of the form $log_b(a)$ as $b$ tends to zero and $a$ either tends to zero or diverges to infinity. Such limits can be said to be "indeterminate of the form $log_0(0)$" or "indeterminate of the form $log_0(infty)$, but this does not mean that they are equal to either of these expressions (anymore than $lim_{xto 0} x^2/x = 0/0$).

Such limits typically require more careful analysis, again using algebraic tools, L'Hospital's rule and other results from analysis, bounding with estimates, or direct $varepsilon$-$delta$ style computation. Limits involving logarithms are discussed in greater detail in J.G.'s answer.

In short, when we say that "the limit is indeterminate of the form $X$", we are saying that if we try to evaluate the limit by naive substitution, then we get the expression $X$, where $X$ is some undefined expression like $frac{0}{0}$, $log_0(0)$, or $1^infty$. Such limits cannot be evaluated by naive substitution, and require other techniques.

Suppose $f,,gto 1$ as $xto 0$. How can we vary $lim_{xto 0}log_f g$? We can take $f=exp x,,g=exp cx$ to get any value $cinBbb R$ we like, or to get $pminfty$ we can use $f=exp x^2,,g=exppm 1$.

Suppose $f,,gto 0$ as $xto 0$. How can we vary $lim_{xto 0}log_f g$? We can take $f=x,,g=x^c$ to get any value $cinBbb R^+$ we like, or to get $infty$ we can use $f=x,,g=x^{1/x^2}$. We can't achieve negative limits because $ln f,,ln gto -infty$.

Let $a=log_1(1)$ and suppose $a$ exists, then $$a=log_1(1)iff 1^a=1$$ so $a$ can be any number. Similarly for $b=log_0(0)iff 0^b=0$