I am trying to prove that $e^x$ is a solution to $f'(x)=f(x)$, and there is a point at which I am concerned that there might be circular logic.

Here's what I've got so far: $$f'(x)=f(x)$$ $$\frac{f'(x)}{f(x)}=1$$ $$\int\frac{f'(x)}{f(x)}dx=x+c_0$$ Letting $y=f(x)$ gives $$\int\frac{dy}{y}=x+c_0$$ Which gives $$\ln|y|=x+c_0$$ $$f(x)=c_1e^x$$ QED

The bit I'm concerned about is $\int\frac{dy}{y}=\ln|y|$.

Is there any proof of $$\int\frac{dx}{x}=\ln|x|$$ which doesn't use $\frac{d}{dx}e^x=e^x$?

Edit: If it wasn't clear, I am asking if there is a way that one can prove that $$\int\frac{dx}{x}=\ln|x|$$ without relying on the fact that $\frac{d}{dx}e^x=e^x$?

  • 13
    How do you define $e^x$? Some people define $\ln(x) := \int_1^x \frac{1}{t} \, dt, \; x>0$ and then define $e^x$ as its inverse. In that case, there is no circular reasoning. – MisterRiemann Oct 27 at 21:00
  • 1
    Assuming you have defined $e = \lim_{n \to \infty} (1+1/n)^{n}$, you could differentiate $\ln(x)$ directly and appeal to this limit. See math.stackexchange.com/questions/1341958/… – welshman500 Oct 27 at 21:03
  • 1
    @Sobi Using the definition of $\exp(x)$ to be the inverse of $\log(x)=\int_1^x \frac{1}{t}\,dt$, one would need to still show that $\exp(x)=(e)^x$, which is not directly evident! – Mark Viola Oct 27 at 21:12
  • 4
    @MarkViola If one first define $\log$ from the integral and then $\exp$ as its inverse function, then one can define $a^x=\exp(x\log a)$; in the particular case of $a=e=\exp(1)$, we get $e^x=\exp(x)$. No need to have predefined general exponential functions. – egreg Oct 27 at 21:25
  • 1
    And how does one know that $\exp(x)$ has the properties of an exponential function without showing this from your defining it as the inverse of the integral representation of the logarithm? Of course, this is not difficult to do. My "point of view" is only this: it is not immediately obvious. Does that clarify my comment? – Mark Viola Oct 27 at 23:04
up vote 5 down vote accepted

Here's a way to remove the circularity.

Define $g(x) = \int_1^x dt/t$. Then $$ g(xy) = \int_1^{xy}\frac{dt}{t} = \int_1^x\frac{dt}{t} + \int_x^{xy}\frac{dt}{t} = \int_1^x\frac{dt}{t} + \int_1^y\frac{du}{u} = g(x)+g(y) $$ where $t = xu$ is used for the substitution. This property is unique to logarithm functions, and its base will be the number $e$ such that $g(e) = 1$. So $g(x) = \log_e(x)$, and $g^{-1}(x) = e^x$.

  • just for clarification: are you defining $e$ as the base of the logarithm? – clathratus Oct 27 at 22:14
  • 1
    Yes. The base of any logarithm function is the value it maps to 1. – eyeballfrog Oct 27 at 22:16
  • Okay I think I get it... We define $g(x)$ as the function that satisfies $$g(xy)=g(x)+g(y)$$ Then we note that its a logarithm, then we define $e$ as its base? – clathratus Oct 27 at 22:19
  • 1
    No, $g$ is defined by $g(x) = \int_1^x dt/t$. We then show it has the property $g(xy) = g(x) + g(y)$, which means it must be a logarithm to some base. Then we define $e$ as that base. – eyeballfrog Oct 27 at 22:26
  • Ohh okay. That makes sense. Thanks! – clathratus Oct 27 at 23:02

Assuming the question you are asking is indeed that which you are trying to solve, to show that $e^x$ is a solution to $f' = f$ all you have to do is show that $\frac{d}{dx}e^x = e^x$. Which can be fairly easily done by looking at the series definition of $e^x$ term by term and differentiating.

If you're interested in investigating whether or not $ke^x$ is the only solution to $f' = f$ (which it is), then other steps are necessary.

  • I know how to prove that $De^x=e^x$ with series, but I want to know if solving the ODE is a valid proof – clathratus Oct 27 at 21:09
  • Using the definition of $\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$, one would still need to show that $\exp(x)=(e)^x$, which is not directly evident. – Mark Viola Oct 27 at 21:14
  • Well, the only other step is invoking Cauchy's / Picard's theorem on uniqueness of solution to an ODE. – Federico Poloni Oct 28 at 9:09

The usual proof of this is by setting $g(x)=f(x)e^{-x}$ and prove $g'=0$.

See for instance: Proof that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$

Nevertheless your reasoning is correct, but solving the ODE this way lacks of rigour (dividing by $f(x)$ without discussing the annulation). Instead it should be used as a hint to find that solutions looks like $Ce^x$ then by CL theorem you claim their maximality and uniqueness.

See for instance LinAlgMan's answer here: https://math.stackexchange.com/a/409974/399263

  • I like your approach here. Thank you – clathratus Oct 27 at 22:20

Let's suppose we know nothing about exponential and logarithm. Now we're given the Cauchy problem \begin{cases} f'(x)=f(x)\\[6px] f(0)=1 \end{cases}

Let's suppose a solution exists defined over $\mathbb{R}$. For $y\in\mathbb{R}$, define $f_y(x)=f(x+y)$. Then, differentiating with respect to $x$, $f'_y(x)=f(x+y)=f_y'(x)$ and $f_y(0)=f(y)$. If $f(y)\ne0$, we obtain that $g(x)=f_y(x)/f(y)$ is a solution of the same Cauchy problem, so we conclude that $g=f$, so that $f(x+y)=f(x)f(y)$ (at least, when $f(y)\ne0$).

Consider $Z=\{x\in\mathbb{R}:x>0, f(x)=0\}$ and assume it is not empty. Then, if $z=\inf Z$, we have by continuity that $f(z)=0$. Now $z>0$, because $f(0)=1$, so $f(z/2)\ne0$. But we have just proved that $f(z)=f(z/2+z/2)=f(z/2)^2\ne0$: a contradiction. Similarly, $f$ cannot have a negative zero. Thus $f$ is everywhere positive and strictly increasing.

In particular $f(1)>1$ and from $f(n)=f(1)^n$ we deduce that $$ \lim_{x\to\infty}f(x)=\infty $$ From $f(0)=f(x)f(-x)$, we get therefore that $\lim_{x\to-\infty}f(x)=0$.

The inverse function $l$ of $f$ is thus defined over $(0,\infty)$. For every $x\in(0,\infty)$, we have $f(l(x))=x$, so by differentiating, $$ 1=f'(l(x))l'(x)=f(l(x))l'(x) $$ and so $l'(x)=1/x$. By the fundamental theorem of calculus, since $l(1)=0$, $$ l(x)=\int_1^x\frac{1}{t}\,dt $$ Since $$ l(xy)=\int_1^{xy}\frac{1}{t}\,dt=\int_1^x\frac{1}{t}\,dt+\int_x^{xy}\frac{1}{t}\,dt $$ In the second integral we can do the substitution $t=xu$, getting $$ l(xy)=\int_1^x\frac{1}{t}\,dt+\int_1^{y}\frac{1}{u}\,du=l(x)+l(y) $$ Now we can drop our assumption about the existence of $f$, because we can consider the function $$ \log x=\int_1^x\frac{1}{t}\,dt $$ which satisfies $\log(xy)=\log x+\log y$, is increasing and has $$ \lim_{x\to0}\log x=-\infty,\qquad\lim_{x\to\infty}\log x=\infty $$ by considering that $\log(2^n)=n\log2$ and $\log2>0$.

The inverse function $\exp$ of $\log$ is therefore a solution of our Cauchy problem.

Now it's not difficult to show that every solution of the differential equation $f'(x)=f(x)$ is of the form $f(x)=c\exp x$. Indeed, if $f$ is such a solution, then $$ h(x)=f(x)\exp(-x) $$ is constant, as $$ h'(x)=f'(x)\exp(-x)-f(x)\exp(-x)=0 $$ Thus $c=f(0)$.

Can we say that $\exp x=e^x$? This is not really difficult. Note that $\log'1=1$, so $$ 1=\lim_{n\to\infty}\frac{\log(1+1/n)-\log1}{1/n} $$ From the main property of $\log$, we get that, for integer $n$ $$ \log(x^n)=n\log x $$ Thus we have $$ 1=\lim_{n\to\infty}\log\left(\left(1+\frac{1}{n}\right)^n\right) $$ and, by continuity, $$ \exp1=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e $$ For positive integer $n$, we have $\exp n=\exp(1+1+\dots+1)=(\exp 1)^n=e^n$. For negative $n$, $1=\exp(n-n)=\exp(n)\exp(-n)$, so also in this case $\exp n=e^n$.

If $p$ is an integer and $q$ is a positive integer, we have $$ e^p=\exp p=\exp(q(p/q))=(\exp(p/q))^q $$ and so $\exp(p/q)=e^{p/q}$. Since the functions $x\mapsto\exp x$ and $x\mapsto e^x$ coincide over the rationals, they are the same.


Actually, the functions $\exp$ and $\log$ allow to prove the existence of $q$-th roots. Let's work under the assumption we don't know them, so we don't know the existence of the functions of type $a^x$ (for $a>0$) for lack of tools.

If $a>0$, we note that $a^n=\exp(n\log a)$, so we can define $a^x=\exp(x\log a)$. By the very definition, it follows that $$ (a^{1/n})^n=(\exp(\tfrac{1}{n}\log a))^n=\exp(n\tfrac{1}{n}\log a)=\exp\log a=a $$ so $a^{1/n}$ is the (unique) positive real number whose $n$-th power is $a$.

By defining $$ e=\exp1=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n $$ we have that $e^x=\exp x$ by definition, but also that, for a rational $p/q$ and any positive $a$, $$ (a^{p/q})^q=a^p $$ so our definition of the general exponential function is what we expect it to be.

  • Great answer. Thank you – clathratus Oct 28 at 0:37

This is one way to define the logarithmic functions.

$$\ln x= \int _1 ^x \frac{dt}{t}, \text { for x > 0 } $$

So with this convenience, your proof is OK.

  • 1
    does this definition ensure that $e^x$ and $\ln x$ are inverses? – clathratus Oct 27 at 21:04
  • @clathratus No, it does not. That is why they define $e^x$ to be the inverse of $\ln x $ – Mohammad Riazi-Kermani Oct 27 at 21:06
  • 3
    Using the definition of $\exp(x)$ to be the inverse of $\log(x)=\int_1^x \frac{1}{t}\,dt$, one would need to still show that $\exp(x)=(e)^x$, which is not directly evident! – Mark Viola Oct 27 at 21:11
  • @MarkViola how does one do so (prove that $\exp(x)=e^x$)? – clathratus Oct 27 at 21:14
  • 1
    See THIS and THIS, THIS, and THIS. – Mark Viola Oct 27 at 21:23

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.