Does ${f(x)=\ln(e^{x^2})}$ reduce to ${x^2\ln(e)}$ or ${2x\ln(e)}$? I'm confused with the expression  ${f(x) = \ln(e^{x^2})}$.I know the rule  ${\log_a(x^p) = p\log_a(x)}$. So does the given expression reduce to ${x^2\ln(e)}$ or ${2x\ln(e)}$?
 A: 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 ${x^2\ln(e)=x^2}$
, 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 ${2x\ln(e)=2x}$.
In any case - if you want to clear any ambiguity in expressions - always put brackets. But like I said, convention would dictate that ${e^{x^2}=e^{(x^2)}}$
A: 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:
$$\log_{e}((e^x)^2) = 2\log_{e}(e^x) = 2x \log_{e}(e) = 2x$$
That's different from what what we got above yea?
A: Here you have $\ln(e^p)$ with $p = x^2$, so the correct reduction is $(x^2)\ln(e) = x^2$.
