Show that $f(x) = e^x $ is a function. Can I simply say the following? I feel like I'm either over or under-thinking this, but it makes sense to me since a function has one y for every x and a one-to-one function has the same criteria.
Let $e^a = e^b$
then ln $(e^a) = $ ln$(e^b)$ $\Rightarrow a = b$
Therefore, $e^x$ is a one-to-one function. Since this is a type of function, 
$e^x$ is a function.
 A: You proved the relationship is one-to-one, but did not prove it is a function. The usual definition is that a relation between two sets (domain and codomain, here both $\mathbb{R}$) defines a function if any element of the domain is associated with exactly one element of the codomain. Here, the definition $f(x) = e^x$ associates with $x$ in the domain exactly one number $e^x$ in the codomain, hence $f$ is indeed a function.
As an example, consider the relation $g:\mathbb{R} \to \mathbb{R}$ defined by assigning to $x \in \mathbb{R}$ the $y$ which satisfies the equation $x = y^2$. Clearly then, some elements in $\mathbb{R}$ (negative numbers) are not even mapped by $g$, some others (positive numbers) are mapped to two values at the same time (e.g. $9 \mapsto \{3, -3\}$) and only $0$ is mapped uniquely. Hence, $g$ would not be a function.
As another example, consider the relation $h:\mathbb{R} \to \mathbb{R}$ associating $x \mapsto 1/x$. This is clearly not a function either since $h(0)$ is not defined. But if we restrict $h:\mathbb{R}^+ \to\mathbb{R}$, it becomes a function.
A: Function definitions usually need a specified domain and codomain, as well as the rule by which the elements from the domain correspond to the elements of the codomain. The answer to your question may depend on the specified domain.
For instance, if the domain is $\mathbb N, \mathbb Z,\mathbb Q$ or $\mathbb R$, then the value of the function $f(x)=e^x$ exists due to the properties of real exponents, namely $e\in\mathbb R^+$.
If the domain is $\mathbb C$, then $e^x$ is primarily defined in terms of the limit definition $$e^x:=\lim_{n\to\infty}{(1+\frac xn)^n}$$ or Maclaurin series $$e^x:=\sum_{n=0}^\infty{\frac{x^n}{n!}}=1+x+\frac{x^2}{2}+\cdots$$ both of which are needed to be proven to converge.
Other domains are possible too, such as matrices and finite fields, but I presume your question was about $f:\mathbb R\rightarrow\mathbb R^+$. In this case, $e^x$ is an exponent with a strictly positive base and a real power for each $x$, which is sufficient (Christopher Thomas (1998), ch. 2).
