In my Calculus class I have seen the definition of $\ln(x)$ and $e^x$ like this: First we define $$ \ln(x) = \int_1^x \frac{1}{t}\; dt $$ Then we show that $\ln(x)$ is one-to-one. Then we define the inverse of $\ln(x)$ as $e^x$. And so on.
Is there a meaningful way to define this without using integrals? If so, what is an outline for how this can be done?
Alternative (and equivalent definitions of $e^x$ include:
$$e^x=\lim_{n\to\infty} \left(1+\frac xn\right)^n$$
$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$
$$e^x=\lim_{q\to x\\ q\in\mathbb Q} e^q$$
The last definition may appear circular, but it is not. In that last definition, we first define $$e=\lim_{n\to\infty} \left(1+\frac1n\right)^n$$
Then, for all rational numbers $q\in\mathbb Q$, we define $$e^q=\sqrt[b]{e^a}$$ where $a,b\in\mathbb Z$ are such that $q=\frac ab$.
I actually like to use the definition I propose below as I think it would be the most natural for anyone new to the topic. First define $$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$ Why is this a good definition? For many it is natural because it arises from a real world example using bank interest.
First important question: Does this limit even exist? If so, why? The answer is yes, and it can be shown by showing that the sequence is bounded (to be exact, $1<e<3$) and increasing, and is well discussed in this MSE post. By using a transformation $m=nx$, we can see that $$\exp(x):=e^x=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}=\lim_{m\to\infty}\left(1+\frac{x}{m}\right)^m.$$ From this we can also derive the power series definition of $e^x$. The binomial theorem: $$(a+b)^n=\sum_{k=0}^n {}_{n}\mathrm{C}_k~a^k~b^{n-k}$$ For integer $n$. And even if $n$ is not an integer, if $n$ is sufficiently large we can use the approximation $\lfloor n\rfloor$ which will converge to the same value as $n\to\infty$. Thus, $$\left(1+\frac{x}{n}\right)^n=\left(\frac{x}{n}+1\right)^n=\sum_{k=0}^n\frac{n!}{(n-k)!k!}\frac{x^k}{n^k}$$ Using a product definition of $n!/(n-k)!$, $$\left(\frac{x}{n}+1\right)^n=\sum_{k=0}^n\left[\prod_{j=1}^k(n-k+j)\frac{x^k}{n^k k!}\right]$$ Since we have $n$ multiplied by iteself $k$ times, we can move it into the product and take the limit: $$\exp(x)=e^x=\lim_{n\to\infty}\sum_{k=0}^n\left[\prod_{j=1}^k\left(\frac{n-k+j}{n}\right)\frac{x^k}{k!}\right]=\sum_{k=0}^\infty \frac{x^k}{k!}.$$ Since $e>1$, $e^x$ is strictly increasing $\forall x\in \Bbb{R}$ and thus has a unique inverse, as proved in this MSE post. With this defintion it is easy to show $$\mathrm{D}(\exp)(x)=\exp(x).$$
We can now define the natural log as the inverse of the exponential, that is, $\ln$ is the unique function such that $$\exp(\ln(x))=\ln(\exp(x))=x ~\forall x\in\mathbb{R}.$$ I suppose the only snag is if we have $x\leq 0$ on the L.H.S. But, in that case we can extend the definition of these functions to complex values and still have consistency. From this we can actually derive the integral definition of $\ln$. $$\mathrm{D}(\exp\circ\ln)(x)=\mathrm{D}(\operatorname{id})(x)$$ $$\exp(\ln(x))\mathrm{D}(\ln)(x)=1$$ $$\frac{\mathrm{d}}{\mathrm{d}x}\ln(x)=\frac{1}{x}$$ $$\ln(x)=\int\frac{1}{x}\mathrm{d}x+C$$ Since $\exp(0)=1$, we can conclude $\ln(1)=0$ and therefore our integral can be written as $$\ln(x)=\int_1^x \frac{1}{t}\mathrm{d}t.$$
