Prove/Disprove $f(x)=e^{x}$ is Injective and Surjective Dr. Pinter's "A Book of Abstract Algebra" presents the exercise:

Prove whether function $f:\mathbb{R} \to (0, \infty)$ denoted by $f(x)=e^x$ is or is not (a) injective and (b) surjective.

Claim: $f$ is injective: suppose $f(a)=f(b)$, then
$$e^a=e^b$$
$$\ln(e^{a})=\ln(e^b)$$
$$a=b$$
Thus, the mapping is injective.
Claim: $f$ is surjective
$$f(x)=e^x$$
$$y=e^x$$
$$\ln(y)=\ln(e^x)$$
$$x=\ln(y)$$
And then plug $x$ into $f(x)$.
$$f(\ln(y))=e^{\ln(y)}$$
$$        =y$$
Thus, the mapping is surjective.
 A: You are admitting the existence of $\log$ in order to prove $e^x$ is injective and surjective... this is circular. $\log$ existing assumes that $e^x$ is a bijection.
(I'll work with $e^x=\sum \frac{x^n}{n!}$)
Claim: $e^x$ is injective.
If $x\geq 0$, $e^x$ is clearly positive. If $x<0$, $e^x=\frac{1}{e^{-x}}$, hence also positive. Since $e^x$ is its own derivative, we have that $e^x$ has positive derivative everywhere. Hence, it is injective.
Claim: $e^x$ is surjective.
$e^x \geq 1+x$ for $x \geq 0$. This implies $e^x \rightarrow \infty$ as $x \rightarrow \infty$ and $e^x \rightarrow 0$ as $x \rightarrow -\infty$. By the intermediate value theorem, we have out result. 
A: Here is another way:
$\exp(x) >0$ for all $x$, and since $\exp' = \exp$, we see that $\exp$ is
strictly increasing, hence injective.
Since $\exp(x) \ge 1+x$ for $x \ge 0$, we see that $\lim_{x \to \infty} \exp(x) = \infty$. Since $\exp(-x) = {1 \over \exp(x)}$, we see that
$\lim_{x \to -\infty} \exp(x) = 0$. The intermediate value theorem
shows that $\exp$ is surjective (with range $(0,\infty)$).
A: For injectivity, we know that the range of $f(x) = e^x$ is always positive, so if one assumes that $e^a = e^b \ \text{for} \ a, b \in \mathbb{R}$, then this implies that $e^{a-b} = 1$. This implies that $a - b = 0$, concluding that $a = b$.
