Let's take a different perspective than the other answers have taken so far. You can think of prime factorizations as telling you exactly how the natural numbers $\mathbb{N}$ (here I am excluding zero) behave under multiplication; there are these primes $p$, and every natural number is uniquely a product of a bunch of primes. In abstract language, $\mathbb{N}$ under multiplication is a type of algebraic structure called a commutative monoid, and prime factorization tells us that $\mathbb{N}$ is the free commutative monoid on the primes.

So the primes tell us exactly how natural numbers multiply. One way to ask the question for real numbers is:

How do real numbers multiply?

That is, how can we understand the behavior of, say, the positive real numbers $\mathbb{R}_{+}$ as a commutative monoid?

The answer for the real numbers is quite different compared to the natural numbers: multiplication works the same way as addition! Formally, for any positive real $r > 1$ we can consider the exponential

$$\mathbb{R} \ni x \mapsto r^x \in \mathbb{R}_{+}$$

and this is an isomorphism of commutative monoids; what this means is that the familiar exponential law

$$r^{x + y} = r^x r^y,$$

together with the fact that the exponential has an inverse given by the logarithm, establishes that multiplication and addition of real numbers work exactly the same way; we can pass from addition to multiplication by taking the exponential, and we can pass from multiplication to exponentiation by taking the logarithm $\log_r$. This is a very fundamental fact, and before the advent of calculators you used to have to learn how to use slide rules to take advantage of this, and multiply complicated numbers by adding their logarithms.

So, we could say loosely that exponentials and logarithms establish that every positive real number has a "continuous factorization" into a "continuous product" of copies of any fixed real number $r > 1$, and this factorization is unique if we fix $r$. Taking $r$ to be $e$ would be a particularly nice choice but we don't have to.

Let's take a different perspective than the other answers have taken so far. You can think of prime factorizations as telling you exactly how the natural numbers $\mathbb{N}$ (here I am excluding zero) behave under multiplication; there are these primes $p$, and every natural number is uniquely a product of a bunch of primes. In abstract language, $\mathbb{N}$ under multiplication is a type of algebraic structure called a commutative monoid, and prime factorization tells us that $\mathbb{N}$ is the free commutative monoid on the primes.

So the primes tell us exactly how natural numbers multiply. One way to ask the question for real numbers is:

How do real numbers multiply?

That is, how can we understand the behavior of, say, the positive real numbers $\mathbb{R}_{+}$ as a commutative monoid?

The answer for the real numbers is quite different compared to the natural numbers: multiplication works the same way as addition! Formally, for any positive real $r > 1$ we can consider the exponential

$$\mathbb{R} \ni x \mapsto r^x \in \mathbb{R}_{+}$$

and this is an isomorphism of commutative monoids; what this means is that the familiar exponential law

$$r^{x + y} = r^x r^y,$$

together with the fact that the exponential has an inverse given by the logarithm, establishes that multiplication and addition of real numbers work exactly the same way; we can pass from addition to multiplication by taking the exponential, and we can pass from multiplication to exponentiation by taking the logarithm $\log_r$. This is a very fundamental fact, and before the advent of calculators you used to have to learn how to use slide rules to take advantage of this, and multiply complicated numbers by adding their logarithms.

So, we could say loosely that exponentials and logarithms establish that every positive real number $x$ has a "unique continuous factorization" into a "continuous product" of "$\log_r x$ copies" of any fixed real number $r > 1$. Taking $r$ to be $e$ would be a particularly nice choice but we don't have to.