if you are using $e^{rt}$, then since doubling time is (in your chosen unit of time) equal to $1$, we get $e^{(r)(1)}=2$. Taking logarithms to the base $e$, we get $r=\ln 2$.
It simplifies things to use the doubling time as the unit of time. However, in a more complicated situation, where we have say two kinds of cell, what unit shall we use? In the long run, the conventional units of time are more useful.
So the population at time $x$ is $e^{(\ln 2)(x)}$. Since $e^{\ln 2}=2$, this is just a fancy way of writing $2^x$.
Both models can be considered as continuous models, since $x$ is not restricted to be an integer. They are in fact the same model. And yes, we do get results that are not integers, but that is not important. No mathematical model will represent a complex biological phenomenon exactly. If the numbers involved are large, an "error" of half a cell has no practical significance. Indeed, the error is likely to be far larger than that.
The function $e^t$ has nice technical properties. For example, its derivative is $e^t$. the derivative of $2^t$ is the more messy $(\ln 2)2^t$. But anything one can do by using base $e$ can also be done using base $2$, or $10$. There will be small differences of detail, that's all.
Here is a more general framework for your question. Suppose you have functions $f(x)$ and $g(x)$. You compare their growth rates by looking at $$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$
In your problem, $f(x)=3^x$ and $g(x)=2^x$, and this limit is $\infty$, which is what we mean when we say "$3^x$ grows faster than $2^x$."
Then you ask: why does taking the logarithm not yield the same order? (In other words, you're asking: if $f(x)$ grows faster than $g(x)$, why doesn't $\log f(x)$ grow faster than $\log g(x)$? Or, in yet other words: why don't logs preserve the asymptotic ordering?) The answer is that you must now consider $$\lim_{x\to\infty}\frac{\log f(x)}{\log g(x)}$$
And there is no nice way to relate the first limit to this limit, in general.
But in your special case when $f(x)$ and $g(x)$ are exponentials, taking the log gives first-order polynomials, which have the same growth rate: $$\lim_{x\to\infty}\frac{\log 3^x}{\log 2^x}=\lim_{x\to\infty}\frac{x\log 3}{x\log 2}=\frac{\log 3}{\log 2}$$
(Since the limit is finite and non-zero, the new functions have the same growth rate.)
EDIT
No, pure exponentials are not the only family of functions that has this property. Consider $f(x)=x\cdot 3^x$ and $g(x)=x\cdot 2^x$. These functions have different growth rates, because $$\lim_{x\to\infty}\frac{x\cdot 3^x}{x\cdot 2^x}=\infty$$
But their logs, $\log x+x\log 3$ and $\log x+x\log 2$, have the same growth rates, because
$$\lim_{x\to\infty}\frac{\log x+x\log 3}{\log x+x\log 2}=\lim_{x\to\infty}\frac{\frac{1}{x}+\log 3}{\frac{1}{x}+\log 2}=\frac{\log 3}{\log 2}$$
In fact, as this example suggests, any function that is the product of a bunch of functions all of which have growth rate at at most exponential will work: their logarithms will have the same growth rate. Take $f(x)=(\log_3 x)(x^3-4x^2+1)3^x$ and $g(x)=(\log x)(x^2)2^x$ for example. These functions have different growth rates, again, but their logarithms have the same growth rate. The logarithm turns the multiplicative structure into an additive structure dominated by a first-order polynomial.
Best Answer
We have $n^{n-2}=e^{(\log n)(n-2)}$ while the other is $e^{(\log 2)\frac{n(n-1)}{2}}$.
Note that $\frac{n(n-1)}{2}$ grows faster than $(\log n)(n-2)$. The exponential function dramatically increases the disparities in rate of growth.