# On what cases i get 0 denomiator

1. Feb 13, 2009

### transgalactic

$$f(x)=\frac{1}{1+ln|x|}$$

the ln|x| part could be -1 when x=e^-1
correct??

2. Feb 13, 2009

### arildno

Certainly!

Do you have any other solutions to the equation 1+ln|x|=0?

3. Feb 13, 2009

### transgalactic

no
??

4. Feb 13, 2009

### arildno

Well, if ln|x|=-1, what must |x| equal?

5. Feb 13, 2009

### transgalactic

|x|=e^-1
x=-e^-1
x=e^-1

6. Feb 13, 2009

### tiny-tim

Yes.

7. Feb 13, 2009

### transgalactic

thanks

8. Feb 13, 2009

### HallsofIvy

Staff Emeritus
But Arildno's point is that that is not the only value. As you said, the denominator is 0 for $e^{-1}$ or $-e^{-1}$.