Can $\ln(x)$ be defined without integral calculus? Typically, the natural log function is defined in integral calculus as $\ln(x)=\displaystyle\int_1^x \frac{dt}{t}$. Is it possible to define $\ln(x)$ using only differentiable calculus?
I should specify, I am asking this question while in an analysis class, where we are rigorously defining $\ln(x)$. We have no knowledge of $e^x$ to call $\ln(x)$ its inverse.
 A: There is a way to define $ln(x)$ without using calculus at all.
You can define $ln(x)$ such that $ln(xy) = ln(x) + ln(y)$, and $ln(e) = 1$. From this definition, all of the properties of $ln(x)$ (like the change of basis formula, being the inverse of $e^x$, and that its deravitive is $\frac{1}{x}$) can be derived.
Unless $x$ is a rational power of $e$, this definition alone doesn't help very much in computing the value of $ln(x)$, but it is rigorous.
A: One possible way is to  consider any segment $[c,d]\subset \mathbb{R}$ and $a \in \mathbb{R}, a>1$. As function $f(x)=a^x$ is strictly monotone and continuous, then on segment $[\alpha, \beta],\alpha = a^c, \beta=a^d $ we can define Reverse function $x=f^{-1}(y)$, which we call logarithm $y=\log_a x$.
A: You could define
$$\ln(a)=\lim_{h\to 0}\frac{a^h-1}{h}=\left[\frac{d}{dx}a^x\right]_{x=0}$$
Of course, this requires proving the existence of $\lim_{h\to 0}\frac{a^h-1}{h}$ for all $a>0$. This definition also assumes that you have adequately defined $a^h$ for all real $h$. One way to do this is to define $a^{(\cdot)}$ as the unique function satisfying these three properties:

*

*$a^{(\cdot)}$ is anywhere-continuous (i.e. it is continuous at some point)

*$a^1=a$

*$a^{x+y}=a^x a^y$ for all $x,y \in \mathbb{R}$
The first property is necessary for $a^{(\cdot)}$ to be continuous everywhere.
