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.
On
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 $b=\log_0(0)\iff 0^b=0$
On
Suppose $f,\,g\to 1$ as $x\to 0$. How can we vary $\lim_{x\to 0}\log_f g$? We can take $f=\exp x,\,g=\exp cx$ to get any value $c\in\Bbb R$ we like, or to get $\pm\infty$ we can use $f=\exp x^2,\,g=\exp\pm 1$.
Suppose $f,\,g\to 0$ as $x\to 0$. How can we vary $\lim_{x\to 0}\log_f g$? We can take $f=x,\,g=x^c$ to get any value $c\in\Bbb R^+$ we like, or to get $\infty$ we can use $f=x,\,g=x^{1/x^2}$. We can't achieve negative limits because $\ln f,\,\ln g\to -\infty$.
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 $$ \lim_{x\to 0} \frac{x^2}{x} = \frac{\lim_{x\to 0} x^2}{\lim_{x\to 0} x} = \frac{0}{0}. $$ 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, $$ \lim_{x\to 0} \frac{x^2}{x} = \lim_{x\to 0} x = 0. $$ 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 $\lim_{x\to 0} x^2/x = 0/0$).
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.