Thanks to Kalid Azad's book (betterexplained.com), I understand exponential phenomena better.
$ae^{rt}$ gives the growth of '$a$' after '$t$' unit of times and with a "continuous" growth rate of '$r$' (so if $rt = 1$ like in $e^1$ the formula will output the new value of '$a$' after a $100\%$ of growth during $1$ unit of time '$t$', it can also be thought of as a $50\%$ growth during $2$ periods of times, etc…).
The $\ln(x)$ function outputs the amount of time needed to have a certain growth of the quantity '$1$'. e.g. $\ln(2.71\dots) = 1$ (we need $1$ unit of time to transition from $1$ to $2.71\dots$ with $100\%$ continuous growth).
My definitions are not $100\%$ mathematical and precise but I can't visualize and understand $\exp(x)$ or $\ln(x)$ without them (especially their applications in engineering stuff).
The $\ln(x)$ function is the antiderivative of $\frac1x$ (or $\frac1x$ is the derivative of $\ln(x)$), and my question is what's the link between $\frac1x$ and the time needed to have a continuous growth of a rate "$r$" and during $x$ unit of times. For example, why the derivative of the $\ln(x)$, the function returning the time to achieve $100\%$ growth during a time unit, is the inverse of the time unit $x$. What's the intuitive explanation of $\ln(x)$ being the antiderivative of $\frac1x$ ?
$$e^x = e^1, e^2, e^3, e^4,\dots$$
$$\ln(e^x) = 1, 2, 3, 4,\dots$$
$$\frac1x = \frac1{e^1}, \frac1{e^2}, \frac1{e^3}, \frac1{e^4},\dots$$
Best Answer
This answer expands upon @user2661923's comment. Although it deviates from the spirit of the question, at least we are able to provide a justification for why $\int 1/x\,\mathrm{d}x = \ln x$ if $\ln(e^x)=x$.
Given an injective real function $f$ and its inverse $f^{-1}$, we know, by definition, that $(f^{-1}\circ f)(x) = x$ for all $x$ in the domain of $f$. Defining $y=f(x)$, we take the derivative of $(f^{-1}\circ f)(x) = f^{-1}(y)$ w.r.t. $x$ using the chain rule: $$ {\mathrm{d}\over\mathrm{d}x}f^{-1}(y) = {\mathrm{d}\over\mathrm{d}y}f^{-1}(y) \cdot {\mathrm{d}y\over\mathrm{d}x} = 1. $$ If $f$ has the propriety $f'(x)=f(x)$ as well, then $y'= f(x) = y$, therefore $$ {\mathrm{d}\over\mathrm{d}y}f^{-1}(y)={1\over y}. $$ Integrating both sides w.r.t. $y$ from some constant $a$ to a variable $x$, we have, by the fundamental theorem of calculus, $$ f^{-1}(x) - f^{-1}(a) = \int\limits_a^x{1\over y}\,\mathrm{d}y\\ f^{-1}(x)=\int\limits_a^x{1\over y}\,\mathrm{d}y + f^{-1}(a). $$
Now, it would be very convenient to choose $a$ such that $f^{-1}(a)=0$, if possible, so as to simplify our expression for $f^{-1}$. Recalling that $(f^{-1}\circ f)(x)=x$ by definition, it's clear that we must choose $a = f(0)$, so $$ f^{-1}(x) = \int\limits_{f(0)}^x{1\over y}\,\mathrm{d}y. $$
Finally, assuming it has been proven that $f(x)=e^x$ has the propriety $f'(x)=f(x)$, we can evaluate $f(0)=e^0=1$ and show that $$ \ln x \triangleq f^{-1}(x) = \int\limits_1^x {1\over y}\,\mathrm{d}y, $$ that is, if $e^x$ is its own derivative, then its inverse must equal the integral of $1/y$ w.r.t. $y$ from $1$ to $x$.