TITLE: Proof $e^x = \exp(x)$?
QUESTION [6 upvotes]: Define $$\ln (x) = \int^{x}_{1}\frac{1}{t}$$
Assume I have proven that $\ln x$ is one-to-one and therefore has an inverse $\exp (x)$.
Define $e$ as:
$\ln e = 1$ 
Now, if you have no other notion of exponentials, or logarithms, how could define what $e^x$ means and show that its the inverse of $\ln x$?
You are allowed to assume the logarithmic product and quotient property.
Thanks for the help.

REPLY [2 votes]: Preliminaries:
First, you don't need to assume the product/quotient rule, as they follow from the definition:
$$\ln (xy) = \int^{xy}_{1}\frac{1}{t}=\int^{x}_{1}\frac{1}{t} +\int^{xy}_{x}\frac{1}{t}= \int^{x}_{1}\frac{1}{t} dt +\int^{xy}_{x}\frac{x}{xt}dt$$
Now, if you make the substitution $u=xt$ in the second integral, you get:
$$\ln (xy) =  \int^{x}_{1}\frac{1}{t} dt +\int^{y}_{1}\frac{1}{u}du=\ln(x)+\ln(y)$$
The quotiont rule for the logarithm follows from here, or can be proven the same way.
Second, it follows from the FTC that $\ln'(x)=\frac{1}{x}$.
Finally, the definition of the logarithm makes sense for $x>0$, and $\ln'(x)=\frac{1}{x} >0$ on its domain, which implies that $\ln(x)$ is one to one. It therefore has an inverse.
It can also be proved from the definition that the range of $\ln$ is $\mathbb R$. Therefore it has an inverse $\exp(x) : \mathbb R \to (0, \infty)$, and all these properties follow from your definition, nothing needs to be assumed.
The proof that $\exp(x)=e^x$
Since $\ln(xy)=\ln(x)+\ln(y)$ its inverse function satisfy the functional equation 
$$\exp(x+y)=\exp(x) \exp(y)$$
From here [exactly as in the Cauchy equation], one gets by induction that 
$$\exp(nx)=[\exp(x)]^n$$
for all positive integers, and then for all integers.
From here it also follows that 
$$\exp(\frac{p}{q})=[\exp(1)]^{\frac{p}{q}} \,.$$
As you defined $e=\exp(1)$, this shows that 
$$\exp(r)=e^r \, \forall r \in \mathbb Q \,.$$
Finally, since $e >1$ (which follows from the definition of the logarithm), the function $e^x$ is increasing [note that since you defined $e=\exp(1)$ you should not use the derivative here, as you don't know if your didn't get a different $e$].
Moreover, it follows immediately from the above definition of $\ln(x)$ that $\ln(x)$ is a strictly increasing function, and therefore, so is its inverse.
Now, your claim follows immediately from the following lemma applied for f(x)=e^x, g(x)=\exp(x)$:
Lemma Let $f(x), g(x)$ be two increasing functions. If $f(x)$ is continuous and $f(x)=g(x) \forall x \in \mathbb Q$ then $f=g$.
The proof is easy, for each $x \in \mathbb R$ pick two sequences of rationals, one which is increasing to $x$, and the other which is decreasing to $x$ and use monotony + continuity of $f$.
Bonus To show that your definition of $e$ is the standard one, note that since $\ln'(x)=\frac{1}{x}$ the formula for the derivative of the inverse function yields
$$[e^x]'=e^x$$