# Why the number e(=2.71828) was chosen as the natural base for logarithm functions? [duplicate]

Possible Duplicate:
What's so “natural” about the base of natural logarithms?

Why the number e(=2.71828) was chosen as the natural base for logarithm functions ? Mainly I am interested in knowing why is it called "natural " . The number "2" could instead have been chosen as the most natural base.

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## marked as duplicate by William, martini, ShreevatsaR, draks ..., Asaf KaragilaJul 29 '12 at 8:51

Just to be clear $e$ is not $2.71828$. $e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n$. You can prove that $e$ is irrational. –  William Jul 29 '12 at 7:53
@William for this question let us stick to e being a finite value and lets not get into the infinite sequence . Technically. though you are correct. –  Geek Jul 29 '12 at 7:55
@Geek If you use the finite value of $e$, you lose the "naturalness" of $e$. Read the wikipedia article on $e$: en.wikipedia.org/wiki/E_(mathematical_constant). There is a section about $e$ and continuously compounded interest. –  William Jul 29 '12 at 7:57
@Geek: $e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n$ is also a finite value. :-) –  ShreevatsaR Jul 29 '12 at 8:13
I don't know why no one except me up-voted this question. –  Michael Hardy Dec 9 '12 at 16:06

If you draw the graphs of $y=a^x$ for varying values of $a$, you find that they all pass through the point$(0,1)$ on the $y$-axis. There is exactly one of these curves that passes through that point with a gradient of exactly 1, and that value is obtained by taking $a=2.718281828459 \dots$.
In more analytical terms, this means that this is the value of $a$ which makes the derivative of $a^x$ equal to $a^x$, rather than a constant multiple of $a^x$.
"There is exactly one of these curves that passes through that point with a gradient of exactly 1..." - not so coincidentally, truncating the series for the exponential function at the linear term yields that particular tangent line with a slope of $1$. ;) –  Ｊ. Ｍ. Jul 29 '12 at 9:50
@HenningMakholm: I'm not sure I understand. $a^x = e^{x\ln a} = 1 + (x\ln a) + (x\ln a)^2/2! + \dots$, and the coefficient of the linear term gives the slope of the tangent line at $x = 0$, namely $\ln a$. This is $1$ precisely when $\ln a = 1$, namely $a = e$. And truncating at the linear term does give the tangent line at $x = 0$, namely the line $y = 1 + x\ln a$ as J. M. said, which has a slope of $1$ for $a = e$. –  ShreevatsaR Jul 30 '12 at 8:28