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.

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    $\begingroup$ 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. $\endgroup$ – Kaster Jun 7 '13 at 10:16
  • $\begingroup$ 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. $\endgroup$ – Saurabh Raje Jun 7 '13 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 :)

  • $\begingroup$ How is the denominator zero? $\endgroup$ – Saurabh Raje Jun 7 '13 at 10:33
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    $\begingroup$ @SaurabhRaje what do you think it is? $\endgroup$ – Kaster Jun 7 '13 at 11:07
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    $\begingroup$ @SaurabhRaje $\log_e1=x\implies 1=e^x\implies e^0=e^x\implies x=0$ $\endgroup$ – iostream007 Jun 7 '13 at 11:15
  • $\begingroup$ Oops.....sorry, I get the point. $\endgroup$ – Saurabh Raje Jun 8 '13 at 10:02

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.


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 ($\mathbb{C}$).

  • $\begingroup$ So, moral of the story is that $\log 1 \ base (1) \not= 1$?? $\endgroup$ – Saurabh Raje Jun 8 '13 at 10:00
  • $\begingroup$ @SaurabhRaje That's correct -- your teacher is wrong. $\endgroup$ – 6005 Jun 8 '13 at 14:09

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



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.


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


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