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Why is $e$ used as a base of natural logarithms everywhere?

Is the origin from the fact that exponential is the only function with the unique property of its differential and integral same and that too equal to the function?

Is there any other function like this?

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5 Answers 5

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The derivative of $\log_ax$ is $${1\over x\log_ea}$$ so even if you try using some base other than $e$ you find base $e$ coming around to bite you. And if you do use base $e$ then the formula simplifies to $1/x$, which is Nature's way of telling you to use base $e$.

And, yes, this is just another way of saying that, up to comstant multiples, $e^x$ is the only function equal to its derivative.

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$f(x)=e^x$ is the only function, up to a constant factor, that has the property $$ \frac{\mathrm{d}}{\mathrm{d}x}f(x)=f(x) $$ Since $e=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n$, it follows that $$ \begin{align} e^x &=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^{nx}\\ &=\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n\\ \end{align} $$ If we accept the exchange of the derivative with the limit, we get that $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}e^x &=\frac{\mathrm{d}}{\mathrm{d}x}\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n\\ &=\lim\limits_{n\to\infty}\frac{n}{n}\left(1+\frac{x}{n}\right)^{n-1}\\ &=\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^{-1}\\[9pt] &=e^x\cdot1 \end{align} $$ Suppose that $\dfrac{\mathrm{d}}{\mathrm{d}x}f(x)=f(x)$, then using the quotient rule $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{e^{x}} &=\frac{e^x\frac{\mathrm{d}}{\mathrm{d}x}f(x)-f(x)\frac{\mathrm{d}}{\mathrm{d}x}e^x}{e^{2x}}\\ &=\frac{e^xf(x)-f(x)e^x}{e^{2x}}\\[6pt] &=0 \end{align} $$ Thus, by the Mean Value Theorem, $\dfrac{f(x)}{e^{x}}$ is a constant. Therefore, $$ f(x)=ce^x $$

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  • $\begingroup$ What property did you use to conclude that $\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{nx}=\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n$ ? $\endgroup$ Jun 14, 2020 at 23:07
  • $\begingroup$ Substitution $n\mapsto nx$. If you are worried about $n\in\mathbb{Z}$, take a look at this answer. $\endgroup$
    – robjohn
    Jun 14, 2020 at 23:34
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The natural logarithm has that name because it is the most natural choice for mathematical purposes as Gerry points out.

Also, we actually only need one logarithm because they are all the same up to a multiplicative constant. That is, given $a>0$ ($a\ne1$) we have $$\log_ax=\log_a(e^{\log_ex})=(\log_e x)\cdot(\log_a e) = C_a \cdot\log_e x.$$


In computer science it is common to use the (less natural) logarithm $\log_2$, while others believe the $\log_{10}$ is the most natural logarithm.

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  • $\begingroup$ This should be made clearer - a learner could be forgiven for thinking that only the natural logarithm has this property. As well as the above derivation it can also be proven using the change of base rule. $\endgroup$
    – user301988
    Apr 30, 2016 at 3:14
  • $\begingroup$ @selfawareuser What part is unclear in your opinion? Note that this is to be with the answer of Gerry Myerson. $\endgroup$ May 7, 2016 at 21:49
  • $\begingroup$ I meant your point should be made clearer elsewhere. $\endgroup$
    – user301988
    May 7, 2016 at 23:51
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Another interesting to note is that if you solve the differential equation $\dfrac{d}{dx}f(x) = f(x)$, you get $\dfrac{\dfrac{d}{dx}[f(x)]}{f(x)}=1$, or $\int {\dfrac{\dfrac{d}{dx}[f(x)]}{f(x)} \text{ d}x}= \int{1 \text{ d}x}$ to get $\text{ln}(|f(x)|) = x + C$, or $|f(x)| = e^{x+C} = C_{1}e^{x} \implies f(x) = C_{1}e^{x}$.

Financial and actuarial fields use powers of $e$, say $e^{\delta t}$ to represent continuous dividend and compounding rates. Actually, these are defined by the $\dfrac{\dfrac{d}{dx}[f(x)]}{f(x)}$ expression.

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I should mention that the use of the base-$e$ logarithm gives rise to some elegant inequalities, like: $$\frac11>\ln\frac21>\frac12>\ln\frac32>\frac13>\ln\frac43>\dotsb$$

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  • $\begingroup$ Note that $\ln x$ is positive iff $x>1$, which is why we have $\ln\frac43$ and things like that rather than $\ln\frac13$. Also, sorry for bringing back a post from 2013. $\endgroup$ Dec 7, 2014 at 2:52

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