$$\newcommand{\RR}{\mathbb{R}}$$
Here is a functional characterization of exponentials; we don’t specify a formula for how to compute the function but instead we specify how the function ought to behave.
(Exponential Functions)
An exponential function is a continuous nonconstant function $E\colon\mathbb{R}\to\mathbb{R}$ satisfying the law of exponents
$$E(x+y)=E(x)E(y)$$
for all $x,y\in\mathbb{R}$.
This note records several properties that follow directly from this functional equation (which prove useful in Part II and III, discussing differentiation and integration). This is material preparing for real analysis, where I am teaching the exponential logarithm and trigonometric functions from a functional-characterization perspective this year.
Positivity
(Exponentials are Nonzero)
If $E$ is an exponential function then $E(x)\neq 0$ for all $x$.
Let $E$ be an exponential function and assume there is some $z\in\RR$ such that $E(z)=0$. Then for any $x\in\RR$ we may write $x=x-z+z=(x-z)+z=y+z$ for $y=x-z\in\RR$. Evaluating $E(x)$ using the law of exponents,
$$E(x)=E(y+z)=E(y)E(z)=E(y)\cdot 0 =0$$
Thus $E$ is constantly zero, which is a contradiction as we assumed $E$ was a nonconstant solution to the Law of Exponents.
(E(0)=1)
If $E$ is any exponential function, then $E(0)=1$.
Since $0=0+0$ apply the law of exponents:
$$E(0)=E(0)E(0)$$
Since we know $E(0)$ to be nonzero we can divide through by it, yielding
$$1=E(0)$$
Because exponentials are continuous (by definition), are nonzero (the proposition above) and return the positive number $1$ at $x=0$ (the other proposition above), the intermediate value theorem ensures exponentials can never have negative number outputs.
Exponential functions are positive.
Uniqueness
(Exponential on Natural Numbers)
Prove that if $E$ is an exponential function, $x\in\RR$ and $n\in\mathbb{N}$ then
$$E\left(nx\right)=E(x)^{n}$$
Since $nx=x+x+\cdots+x$, we may inductively apply the law of exponents:
$$E\left(\underbrace{x+x+\cdots+x}_n\right)=\underbrace{E(x)E(x)\cdots E(x)}_n=E(x)^n$$
(Exponential on Integers)
Prove that if $E$ is an exponential function, $x\in\RR$ and $j\in\mathbb{Z}$ then
$$E\left(jx\right)=E(x)^{j}$$
If $j$ is positive, the result is already proven for natural numbers above. If $j=0$ then the claim reduces to $E(0)=1$, also already proven. So, it remains to consider only the case $j=-n$ for $n\in\mathbb{N}$.
Here, we apply the law of exponents to $n-n=0$ to get
$$1=E(0)=E(nx-nx)=E(nx)E(-nx)$$
Thus $E(-nx)=1/E(nx)$, and we know $E(nx)=E(x)^n$ from the natural number case. Putting this together with the definition of negative exponents,
$$E(jx):=E(-nx)=\frac{1}{E(x)^n}=E(x)^{-n}=E(x)^j$$
(Exponential on Rationals)
Prove that if $E$ is an exponential function, $x\in\RR$ and $r\in\mathbb{Q}$ then
$$E\left(xr\right)=E(x)^{r}$$
Let $r=p/q$ for integers $p,q$. We can assume without loss of generality that $p\neq 0$ (as we’ve covered this case) and $q > 1$ ($q=0$ is not allowed, and we can always collect negative signs into the numerator, and $q=1$ is the integer case, also already covered.)
We proceed in two steps: first we look at fractions of the form $1/n$, so we must evaluate $E(\frac{1}{n}x)$. Here we apply the law of exponents to the identity $1=\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}$:
$$
\begin{align}
E(x)&=E\left(\underbrace{\tfrac{x}{n}+\tfrac{x}{n}+\cdots+\tfrac{x}{n}}_n\right)\
&=\underbrace{E\left(\tfrac{x}{n}\right)E\left(\tfrac{x}{n}\right)\cdots E\left(\tfrac{x}{n}\right)}_n\
&=E\left(\tfrac{x}{n}\right)^n
\end{align}
$$
Since $E(x)$, $E(x/n)$ are known to be positive, we can uniquely take the $n^{th}$ root of both sides to get
$$E(x)^{\frac{1}{n}}=E\left(\tfrac{1}{n}x\right)$$
Now let $r=p/q$ be an arbitrary fraction, and consider $E(\tfrac{p}{q}x)$. We compute as
$$
\begin{align}
E\left(\tfrac{p}{q}x\right) &= E\left( p\left[\tfrac{1}{q}x\right]\right)\
&=E\left(\tfrac{1}{q}x\right)^p\
&= \left(E\left(x\right)^p\right)^{\frac{1}{q}}\
&= E(x)^\frac{p}{q}
\end{align}
$$
Thus $E(rx)=E(x)^r$ as claimed.
(Exponentials Agreeing at a Point)
If $E,F$ are exponentials which take the same value at any nonzero point $c$, then they agree on the entire real line.
Assume $E(c)=F(c)=a$, and consider the set $S$ of rational multiples of $c$:
$$ S={ c r\mid r\in\mathbb{Q}}$$
This is dense in $\RR$ and for every $s\in S$ we can calculate $E(s)$ and $F(s)$ using the previous proposition
$$E(s)=E(rc)=E(c)^r=a^r$$
$$F(s)=F(rc)=F(c)^r=a^r$$
Thus $E(s)=F(s)$ on all of $S$, so the functions agree on a dense set, and by continuity, agree on all of $\RR$.
(Base of an Exponential)
If $E$ is an exponential function, its base is defined as the value $a=E(1)$.
(Uniqueness with Given Base)
If two exponentials have the same base, they are equal on the entire real line.
By defintion they agree at $1$, so apply the previous proposition.
Monotonicity & Convexity
If $E$ is an exponential, then restricted to the positive reals $E(x)$ is either (i) always greater than 1, or (ii) always less than 1.
Assume there are two positive numbers $a,b$ with $E(a) > 1$ and $E(b) < 1$. Then by the intermediate value theorem there must be a point $c\in (a,b)$ where $E(c)=1$.
Now consider the set of rational multiples of $c$:
$$ S={ c r\mid r\in\mathbb{Q}}$$
This is dense in $\RR$ and for every element $s\in S$ we see
$$E(s)=E(rc)=E(c)^r=1^r=1$$
Thus $E$ is constantly equal to 1 on the dense set $S$, and so by continuity is constantly equal to $1$ on all of $\RR$. This contradicts the definition of exponentials, but even more immediately contradicts the premise of this proposition, that $E(a) >1$ and $E(b) <1$.
(Exponentials are Monotone)
If $E$ is an exponential function then $E$ is strictly monotone: it is monotone increasing if $E(x) > 1$ on the positive reals, and decreasing if $E(x) < 1$ there.
Without loss of generality we work with the case that $E > 1$ on the positives.
Let $x < y$ be arbitrary real numbers, we wish to show that $E(x) < E(y)$. Since $y-x > 0$ we know $E(y-x) > 1$ and by the law of exponents
$$E(y)= E(y-x+x)=E(y-x)E(x) > 1 E(x)$$
So, $E(y) > E(x)$ as required.
Next we look at convexity
(Convex Function)
A function $f$ is convex on an interval if the secant line between any two points in the interval lies strictly above the graph of $f$.
(Exponentials are Convex)
Let $E$ be an exponential. Then $E$ is convex on all of $\mathbb{R}$.
