Is $\log_{1}{(1)}$ undefined or can be any real number. Let me explain
When we ask what is $\frac{0}{0}$. The answer comes out to be undefined. This follows immediately from the definition of division of two number $a,b$ which says that $\frac{a}{b}$ is the unique solution to the equation $x\times b=a$. Now if we put $a=b=0$ i.e. $\frac{0}{0}$ what we are looking for is the unique solution to the equation $x\times 0=0$. At first sight it might seem that answer can be any real number but according to definition answer should be unique and since answer is not coming unique so we say answer is not any number or undefined
Now I come to the question of $log_{1}{(1)}$ . When we ask what is $log_{a}{(b)}$ what we mean is the solution(not sure is it unique or not) to the equation $a^x=b$ and if $a=b=1$ then solution(not sure unique or not) to equation $1^x=1$ . Now the whole thing depends upon my being unsure about whether the definition includes unique solution(to $a^x=b$) in case of logarithms also as it is in the case of division.
It’s very clear that if definition includes the “unique” solution to $a^x=b$ (as in the case of division) then answer to $log_{1}{(1)}$ will be undefined and if this is not the case then the answer will be any real number.
So does we include the word “unique” or not and thus answer is undefined or any real number.
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$\begingroup$$f(x) = \log_b(x)$ is defined to be the inverse of the exponential function $g(x) =b^x$. Since $x \mapsto 1^x$ is not injective, it does not have an inverse. Hence $\log_{1}(x)$ is not well-defined.
$\endgroup$ $\begingroup$The expression $\log_1(1)$ is the "logarithmic equivalent" of $0/0$.
You may have heard of the rule$$\log_x(y)=\frac{\log_a(x)}{\log_a(y)}$$for any $a,x,y$ for which all of the expressions are defined. If you set $x,y=1$ and $a$ to be some non-one value, you get that $\log_1(1)$ should be$$\frac{\log_a(1)}{\log_a(1)}=\frac 00.$$When we define $\log_x(y)$ as "the unique number $a$ so that $x^a=y$," we run into the error you mentioned where $1^a=1$ for all real $a$. Similarly, when we define $x/y$ as "the unique number $a$ so that $ay=x$," we run into the error where $0\cdot a = 0$ for all $a$. It's the same issue, just in the exponent.
As such, we say that $\log_1(1)$, like $0/0$, is undefined.
$\endgroup$ $\begingroup$I don’t think logarithms are defined for base 1.
Specifically, since $\log_a{b}=\frac{1}{\log_b{a}}$,
we can consider $\log_x{1}=0$, using the above property above: $\log_1{x}=\frac{1}{\log_x{1}}=\frac{1}{0}$
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