# log base 1 of 1

What is $\log(1)$ to the base of $1$? My teacher says it is $1$. I beg to differ, I think it can be all real numbers! i.e., $1^x = 1$, where $x\in \mathbb{R}$.

So I was wondering where I have gone wrong.

• Although it's the matter of convention, but usually $\log_b a$ is not defined for $b = 1$. But if you want to find a formal solution of $1^x = 1$, then you're right. Any complex power of $1$ is $1$. So the solution is $x \in \mathbb C$, or $x \in \mathbb R$ if you're working with real numbers. Commented Jun 7, 2013 at 10:16
• Yeah but my original question was in log form. I converted it to exp form to make it more intuitive. I would appreciate if ur answers r in log formats then it would rule out any problems at that stage of conversion. Commented Jun 7, 2013 at 10:36

The reason why it is not convenient to define $\log$ for the base of $1$ is simple:

$$\log_11=\frac{\log_e 1}{\log_e 1}$$

But the denominator is $0$ and thus the division doesn't make any sense unless we're working with limits :)

• How is the denominator zero? Commented Jun 7, 2013 at 10:33
• @SaurabhRaje what do you think it is? Commented Jun 7, 2013 at 11:07
• @SaurabhRaje $\log_e1=x\implies 1=e^x\implies e^0=e^x\implies x=0$ Commented Jun 7, 2013 at 11:15
• Oops.....sorry, I get the point. Commented Jun 8, 2013 at 10:02
• You are forgetting something. Here is Euler's identity. $e^{i𝜋}+1=0$. So $(e^{i𝜋})^2=1$. $i2𝜋$ is a natural logarithm of $1$. Commented Feb 9 at 6:45

What is $\dfrac00$? What number must $x$ be if $0\cdot x=0$? It can be any number.

What is $\log_1 1$? What number must $x$ be if $1^x=1$? It can be any number.

Hence these expressions are undefined.

What is $\lim\limits_{x\to a}\dfrac{f(x)}{g(x)}$ if $\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a} g(x)=0$? In some cases it's $6$. It depends on which functions $f$ and $g$ are. It can be any number or $\infty$ or $-\infty$. But it's not always undefined. In many cases it's defined and equal to a particular number. For that reason $\dfrac00$ is an indeterminate form.

What is $\lim\limits_{x\to a}\log_{f(x)}g(x)$ if $\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a} g(x)=1$? Again this depends on which functions $f$ and $g$ are. In many cases it's a specific number. This is also an indeterminate form.

• 0/0, log1(1) are the same, and tan(90), sin(90)/cos(90) and a/0 are the same. They claim it is undefined due to division but I beg to differ. It is well defined. In the slope formula (y2-y1)/(x2-x1) = dy/dx = sin(t)/cos(t) = tan(t) it shows up when t = 90 or the slope approaches (+/-)infinity. For me the wording of undefined is wrong. The answer could be 0, could be 1, could be +/-infinity. Without more information and proper context we don't know thus the correct answer is that x/0 for all x is Ambiguous not Undefined! Undefined to me means there is no definition. Ambiguous is more accurate! Commented Oct 21, 2021 at 3:20

If we defined $\log_1 1$, we would want it to satisfy the basic properties that log satisfies. One of these properties is

$$a^{\log_a b} = b$$

Well, this is bad, because setting $a = 1$, we find that $1^{\log_1 b} = 1^{\text{stuff}} = 1$, so the equation works only when $b = 1$. But suppose we ignore this property. There are still other properties of $\log$ we can't satisfy, like the change of base formula: $$\log_a {b} = \frac{\log_c b}{\log_c a}$$

As Sarunas nicely observes, setting $a = b = 1$ gives $\log_1 1 = \frac00$, which is bad. But suppose we ignore this problem as well. Then another property of logs is $$\log_b x + \log_b y = \log_b xy$$

Well, $b = x = 1$ gives $$\log_1 1 + \log_1 y = \log_1 y \implies \log_1 1 = 0$$

which suggests that $\log_1 1 = 0$. But there are still other properties: $$\log_a a^b = b$$ Well, setting $a = 1$, we get $\log_1 1 = b$, and this must be true for any $b$. So we have yet another problem.

We could go on like this for ages, but hopefully you get the idea. While you can define $\log_1 1$, you're going to run into problems because virtually all of the properties of logs are no longer satisfied in the way you want.

It's worth noting that in complex analysis, $\log$ in general has to be a multivalued function, i.e. since there are multiple solutions in $x$ to $a^x = b$, $\log_a b$ has multiple values. From this standpoint, it makes a lot of sense to define $\log_1 1$ as the set of all complex numbers ($\mathbb{C}$).

• So, moral of the story is that $\log 1 \ base (1) \not= 1$?? Commented Jun 8, 2013 at 10:00
• @SaurabhRaje That's correct -- your teacher is wrong. Commented Jun 8, 2013 at 14:09
• Here is something interesting. 1 would have an uncountably infinite number of laogarithms to ther base 1, while other nonzero complex numbers only have a countably infinite number of logarithms to the base 1. Commented Feb 9 at 6:47

Well the question itself is wrong at first place, because while making question you need to take care of domain of function, which in this can never be 1.

By definition of logarithmic function, we know that base of logarithmic function is a positive number excluding x =1. x>0,x≠1

So, for f(x)=logx1
x≠1

Source

So what you're saying is completely valid, $1^x = 1$ is an equation for which the solutions are defined by the set $\mathbb{R}$. However the function $\log_b : \mathbb{R}^+ \rightarrow \mathbb{R}$ isn't defined for $\log_1(1)$, as the log function is only defined to return a single real number. What you're suggesting requires that the definition needs to be $\log_b : \mathbb{R}^+ \rightarrow \{a : b^x = a \}$. Changing this definition then results in over complication and doesn't serve the use the old one does as being a component in other real valued functions, at the end of the day this is just a matter of formalisms.

$$1$$ has an uncountably infinite number of logarithms to the base $$1$$. For any nonzero complex number $$c$$, $$1$$ is a $$c$$th power of $$1$$, and the set of complex numbers is uncountably infinite.

Contrast this with complex numbers other than 1. If $$z$$ is a nonzero integer and $$i^2=-1$$, then the logarithms of $$c$$ to the base $$1$$ take the form $$\frac {-i\ln(c)}{2𝜋z}$$. As the nonzero integers are on ly a countably infinite set, there are only a countably infinite number of logarithms to the base 1 for each $$c$$.

log 1 to the base 1 is "0". Because log a base a is 1"where a not equal 0 and a not equal 1" therefore log 1 base 1 is 0

• $1$ has an uncountably infinite number of logarithms to the base 1 among both real numbers and complex numbers. Commented Feb 9 at 18:34