# Log Base 1

1. Aug 7, 2008

### JinM

I know that logarithms to the base 1 is undefined, due to the reason that:

$\log_1{x} = \frac{\log_a{x}}{\log_a{1}}$

And this leads to divison by zero, which is undefined.

There was a question in one of my textbooks that asked describe the graph that results if $y = \log_1{x}$. Is such a graph even possible?

If I switch this logarithm to exponential form I would get:

$1^y = x$

Now, is it possible that the graph could be y = 1 and x = 1? Since $1^y = x^1$, $y = 1$ and $x = 1$.

Or is it the point of intersection of these two lines? If not, what is it?

Thanks.

Last edited: Aug 7, 2008
2. Aug 7, 2008

### HallsofIvy

Well, not "the" graph. You have two "functions" there and so two graphs. Yes, if y= log1(x) then x= 1y= 1 for all x. The reason I put "functions" in quotes above is that this is not actually a function: strictly speaking, log1(x) is not a function. But it is a relation and its graph is the vertical straight line x= 1.

y= 1x= 1 is a function: its graph is the horizontal straight line y= 1.

3. Aug 7, 2008

### snipez90

Well not even going into division by zero when considering a change of base, you should recognize that 1 raised to a power won't get you x, a variable.

1 raised to any power, whether fractional, negative, or worse, is just 1. So the graph will look like x = 1.

4. Nov 20, 2008

### 5inister

Wouldn't the expression y=log11 (or y=log1(x);x=1 for all x) be undifned too as logn1 is always zero; So y=log11 has both zero and one as an answer?