1. ## Confirmation needed

So I'm just about done with my latest assignment, however there's this little vermin impeding my way towards a perfect score. I think I may have nailed the answer, but given that it's my last available attempt at submitting a response, I want to be extra sure.

Some background information before the actual question:

The function...

The expression $f(x)$ is defined exactly on the set...

The expression $f(x)$ is equal to zero if $x$ is equal to...

The set of all real numbers $x$ for which $f(x)$ is defined and non-zero is...

And so, here is what's currently holding me back...

By analyzing the sign of $f(x)$ on the above open intervals, solve the inequality...

... expressing answer in interval notation.

Is it correct?

2. yes.

3. Originally Posted by skeeter
yes.
Why thank you. The answer was submitted and it indeed was correct.

There's one little thing picking my interest though, why exactly can the function not be ± $e$^(8/3)? In other words, $x=0$?

Is there some $Ln$ rule I'm forgetting?

4. Originally Posted by Desperation
Why thank you. The answer was submitted and it indeed was correct.

There's one little thing picking my interest though, why exactly can the function not be ± $e$^(8/3)? In other words, $x=0$?

Is there some $Ln$ rule I'm forgetting?
If I interpret you correctly ...

x cannot equal 0 because $\ln(0)$ is undefined and x = 0 would also make the denominator of the function 0 ... undefined again.

the function equals 0 at $x = \pm e^{\frac{8}{3}}$ ... that's fine, but the inequality in question only asked for those values where the function was greater than 0 ... not 0.

5. Originally Posted by skeeter
If I interpret you correctly ...

x cannot equal 0 because $\ln(0)$ is undefined and x = 0 would also make the denominator of the function 0 ... undefined again.

the function equals 0 at $x = \pm e^{\frac{8}{3}}$ ... that's fine, but the inequality in question only asked for those values where the function was greater than 0 ... not 0.
I was referring to the set of all real numbers $x$ for which $f(x)$ is defined and non-zero; I was wondering why it excluded ± $e$^(8/3).