Yes, there is, but it's a long chain.
The first step is to realize that you don't actually have, from basic arithmetic, a definition of $a^y$ unless $y$ is rational.
So to extend $$x \mapsto a^x$$ to a function defined on the reals, you need more than your basic arithmetic.
Pretending that you know the definition, you can work out that the derivative of this function (if it exists) must have the form
$$
x \mapsto C a^x
$$
for some constant $C$ (which depends on $a$). Then you can even (via the intermediate value theorem, for instance) reason that for some number $a$, the value of $C$ must be $1$, and call this number $e$. From your question, it appears that you "get" at least this step.
To summarize: if there's a consistent definition of $a^x$ for real $x$ rather than just rational $x$, and if the function $x \mapsto a^x$ is differentiable, then there's a number $e$ with $x \mapsto e^x$ differentiable, and whose derivative is $x \mapsto e^x$.
Now you go a step further, and note that (by continuity), the function $x \mapsto e^x$ must be monotone, and surjective onto the positive reals. So we have (if our original assumption is correct) a function
$$
\exp : \Bbb R \to \Bbb R^{+}
$$
that's bijective and differentiable, and whose derivative is nowhere zero.
You can then say "If all this conjectural stuff is true, then "exp" must have an inverse function!" and you give it name, $\log$. And the inverse function theorem tells you that
$$
\log'(x) = \frac{1}{x}.
$$
And a little fiddling shows you that because $exp(0) = 1$, $\log(1) = 0$.
And that is your real starting point. You ask yourself, "Is there any function $L$ with the property that $L(1) = 0$ and $L'(x) = 1/x$?"
And the answer is, "Yes! I can't write down an algebraic formula for it, but the fundamental theorem of calculus proves it must exist, and I can write down a calculus formula for it, namely:
$$
L(x) = \int_1^x \frac{1}{x} ~ dx."
$$
Then you spend a while proving various facts about $L$ (basically, that it behaves just the way you'd expect $\ln$ to behave), and you call it $\ln$. And then (this is the clever part), you define $\exp$ to be the inverse of $\ln$.
Now you do some more work -- quite a lot more -- to show that for every rational $r$, you have $exp(r) = e^r$ (where the right hand side is interpreted in the sense you already know for rules of exponents).
And finally, you conclude that $r \mapsto e^r$ is a continuous function on the rationals, and therefore has at most one continuous extension to the reals, which must therefore be the continuous function $\exp$ that you've built.
At this point, all the familiar rules for $\ln$ and $\exp$ are available to you.
So you look at
$$
S = \lim_{n \to \infty} (1 + \frac{x}{n})^n
$$
and by taking $\ln$ of both sides, you find that $\ln S = x$, hence that $S = e^x$. This requires the theorem that "limits commute with application of continuous functions", carefully applied, but nothing special about logs or exponentials.
In short: yes, they're tied together, but it's by a long chain of reasoning.
For all the details, see Calculus, by Michael Spivak, among others.
