Does ${f(x)=ln(e^{x^2})}$ reduce to ${x^2ln(e)}$ or ${2xln(e)}$?

It depends whether by ${e^{x^2}}$ you mean

$${e^{(x^2)}}$$

or

$${(e^{x})^2}$$

If you intend to say the first, then the answer reduces to ${x^2ln(e)=x^2}$ , and by convention by ${e^{x^2}}$ we usually are referring to ${e^{(x^2)}}$.

If you are referring to the second, the answer indeed reduces to ${2xln(e)=2x}$.

In any case - if you want to clear any ambiguity in expressions - always put brackets. But like I said, convention would dictate that ${e^{x^2}=e^{(x^2)}}$

Here you have $ln(e^p)$ with $p = x^2$, so the correct reduction is $(x^2)ln(e) = x^2$.

From what you said, you know that:

$$log_{a}(u^p) = p cdot log_{a}(u)$$

whenever $a,u,p$ take on values that make sense. In this instance, let $u = e$ and $p = x^2$. Then, we have:

$$log_{a}(e^{x^2}) = x^2 cdot log_{a}(e)$$

Does that make sense? Since $a = e$ in this particular instance, it follows that the above can just be simplified to $x^2$.

Edit:

Okay, so I wasn't entirely sure what you were talking about initially because $e^{x^2}$, by itself, looks very different from $(e^x)^2$. If $u = e^x$ and $p = 2$, then we would have:

$$log_{e}((e^x)^2) = 2log_{e}(e^x) = 2x log_{e}(e) = 2x$$

That's different from what what we got above yea?