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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 $\frac1x$ ?

$$e^x = e^1, e^2, e^3, e^4,\dots$$

$$\ln(e^x) = 1, 2, 3, 4,\dots$$

$$\frac1x = \frac1{e^1}, \frac1{e^2}, \frac1{e^3}, \frac1{e^4},\dots$$

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  • $\begingroup$ Please use LaTeX and Mathjax formatting for future posts. $\endgroup$ Commented Jul 16, 2022 at 19:13
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    $\begingroup$ Posting this as a comment, rather than an answer, because my response is very informal, and not backed up by a solid understanding of the topic. The function $f(x) = e^x$ has the property that it is equal to its own derivative. That is, $f'(x) = f(x).$ Let $g(x) = \ln(x)$. Then $g(x)$ is the inverse function to $f(x)$. There is a principle in Real Analysis that when any $g(x)$ is the inverse of any $f(x)$, and when $f'(x_0) \neq 0$, then $g'(x_0) = \dfrac{1}{f'(x_0)}.$ ...see next comment $\endgroup$ Commented Jul 16, 2022 at 19:40
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    $\begingroup$ So, letting $y = e^x$, you have that $g(y) = x.$ Then $g'(y) = \dfrac{1}{f'(x)} = \dfrac{1}{f(x)} = \dfrac{1}{y}.$ Thus, the derivative of the function $g(y) = \ln(y)$ is $g'(y) = \dfrac{1}{y}.$ $\endgroup$ Commented Jul 16, 2022 at 19:43
  • $\begingroup$ It alludes to how $\sin(t)$ gives the arc length of a circle given an angle $t$, $\arcsin(x)$ gives the angle given the arc length x, and $\arcsin’(x)$ is the integrand. $\arcsin(x)$ can be put in terms of $\ln(x)$ while $\sin(x)$ can be put in terms of $e^x$. As for the intuitive explanation, $\int x^{-1} dx=\lim_{p\to0} \frac{x^p}p=\ln(x)$ $\endgroup$ Commented Jul 16, 2022 at 22:54

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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 $x$.

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  • $\begingroup$ and what about x between 0 and 1 (exclusive) $\endgroup$ Commented Aug 28, 2022 at 18:06
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    $\begingroup$ By the usual proprieties of integrals, $$\ln x = -\int\limits_x^1{1\over y}\ \mathrm dy.$$ By using the substitution $u = t/x$, we find that $$\int\limits_x^1{1\over y}\ \mathrm dy = \int\limits_1^{1/x}{1\over u}\ \mathrm du = \ln {1\over x} = -\ln x$$ $\endgroup$ Commented Aug 29, 2022 at 4:26
  • $\begingroup$ Correction: use the substitution $u=y/x$. $\endgroup$ Commented Aug 29, 2022 at 7:00
  • $\begingroup$ Thus, it's shown how $\ln x$ and $\ln (1/x)$ are related and we can always refer to our integral definition because if $0<x<1$, then $1/x>1$. $\endgroup$ Commented Aug 29, 2022 at 7:01
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@user2661923 comment is a good proof and should be published as an answer; but anyway, here is another one which is much more informal but perhaps more intuitive.

Logarithm has the property that when $x$ is multiplied by a fixed amount, $\log(x)$ gets added a fixed amount: it transforms multiplication into addition. (I use $\log$ instead of $\ln$ here because this is true for a logarithm with any base).

So $\log(x)$ increases by the same amount when $x$ goes from $1$ to $10$, as when $x$ goes from $10$ to $100$. So the average slope must be 10 times less between $10$ and $100$, that it is between $1$ and $10$. We see that when $x$ is multiplied by $a$, the slope gets divided by $a$: the slope must be $c/x$, with $c$ a constant. And the slope is the derivative.

This can be made a little more formal: if $\log$ has a derivative on $x$ and on $y$, these are $\lim_{h \rightarrow 0} \frac {\log(x+h)-\log(x)} h$ and $\lim_{h \rightarrow 0} \frac {\log(y+h)-\log(y)} h$.

We have $\frac {y+h} y = \frac {x + \frac x y h} x$, so using the property that $\log$ transforms multiplication into addition, we have

$\log(y+h) - \log(y) = \log(x + \frac x y h) - \log(x)$, so

$\log'(y) = \lim_{h \rightarrow 0} \frac {\log(y+h)-\log(y)} h = \lim_{h \rightarrow 0} \frac {\log(x + \frac x y h)-\log(x)} h = \lim_{h \rightarrow 0} \frac x y \frac {\log(x + \frac x y h)-\log(x)} {\frac x y h} = \frac x y \log'(x)$

Then call $c = \log'(1)$, $\forall y$ where $\log$ is derivable, $\log'(y) = \frac {\log'(1)} y = \frac c y$

This however does not prove that constant $c$ is $1$ for the logarithm with base $e$.

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  • $\begingroup$ what's "y" and why (y+h)/y=(x+x/y*h)/x $\endgroup$ Commented Jul 16, 2022 at 23:18
  • $\begingroup$ @Aminos $y$ is a positive real number just as $x$. $(x+x/y*h)/x = 1+h/y = (y+h)/y$. $\endgroup$ Commented Jul 16, 2022 at 23:25
  • $\begingroup$ Bear with me please, what's x/y*h is ? $\endgroup$ Commented Jul 17, 2022 at 18:10
  • $\begingroup$ That's $xh/y$. Actually I just copied it from your comment. $\endgroup$ Commented Jul 17, 2022 at 19:36
  • $\begingroup$ I mean what does it mean ? what's the link between x and y. $\endgroup$ Commented Jul 18, 2022 at 9:58

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