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

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.

• Implicitly, you are saying that you do not believe that $\log_0(x)$ and $\log_1(x)$ are undefined. If they are not undefined, what are their definitions? Jan 19, 2019 at 19:40
• @Xander Henderson i am saying division by zero is yet undefined, but it gets more complicated, when numerator is also zero Jan 19, 2019 at 19:43
• No, it really doesn't get any more complicated. $\frac{0}{0}$ is an undefined mathematical expression. Jan 19, 2019 at 19:44
• that's definitely not true, 0/0 and other indeterminants are so important that whole branch of limits grew out of them, they can't be simply undefined Jan 19, 2019 at 19:50
• thnx @Blue for help Jan 19, 2019 at 20:07

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.

• WOW! by the way i am struggling now i think Jan 19, 2019 at 20:15
• so basically it needs more analysis Jan 19, 2019 at 20:16
• all this means 0/0 is undefined Jan 19, 2019 at 20:18
• thnx thnx thnx, really learnt something new Jan 19, 2019 at 20:19
• :) wow really thank you, thnx for breaking the mirage Jan 19, 2019 at 20:19

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 $$\ln f,\,\ln g\to -\infty$$.

• wow! i didnt even noticed that, that's amazing, i didnt thought about there ranges Jan 19, 2019 at 19:57
• @PranshuKandal This limit-theoretic treatment explains all the indeterminate forms.
– J.G.
Jan 19, 2019 at 19:58
• thnx @J.G. for help Jan 19, 2019 at 20:05
• got that as zero to the power of any non-positive number is not defined, log 0 base 0 can only be positive. Jan 19, 2019 at 20:11
• @PranshuKhandal $\log_0(0)$ is undefined. If $f(x) \to 0$ and $g(x) \to 0$, then $\lim \log_{f(x)}(g(x))$ is (naively) "indeterminate of the form $\log_0(0)$", which may be nonnegative, infinite, or undefined. Because the limit is of an indeterminate form when evaluated naively, deeper analysis is required. Jan 19, 2019 at 20:16

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 $$b=\log_0(0)\iff 0^b=0$$

• does it not prove them indeterminant?? and yes, nice name you have got :) Jan 19, 2019 at 19:47
• Indeterminant in the sense that any value works, both $a$ and $b\neq 0$ can take any value. Jan 19, 2019 at 19:51
• can we add them into the list of our fundamental indeterminants Jan 19, 2019 at 19:53
• if not? them why Jan 19, 2019 at 19:53
• i think I've got the answer, but which do i accept, help me out Jan 19, 2019 at 19:55