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If we have to find differential coefficient of $x$ to the power $x$ with respect to $x$ , we take logarithm base $e$ of both sides of $x^x=y$. Why is $e$ selected as base instead of any other number ?

Ovi
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Chaitanya
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    $e$ has properties too numerous to recount. It would be a good idea to go over a wikipedia article on it. As an answer below mentions, it is the one constant that never needs to be multiplied in or divided out during derivation or integration, so it is a natural choice for the base of the "natural" logarithm. – abiessu Aug 14 '13 at 14:48
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    Related: http://math.stackexchange.com/questions/3319/why-is-the-number-e-so-important-in-mathematics, http://math.stackexchange.com/questions/272507/how-would-you-explain-why-e-is-important-and-when-it-applies, http://math.stackexchange.com/questions/88902/what-is-the-constant-e-fundamentally, http://math.stackexchange.com/questions/26037/intuitive-understanding-of-the-constant-e – apnorton Aug 14 '13 at 14:57

2 Answers2

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One reason $e$ is such a prominent base is that $\dfrac{d}{dx} e^x = e^x$, whereas with an arbitrary base $a$, we have that $\dfrac{d}{dx} a^x = \ln(a)a^x$. Using $e$ as the chosen base removes this factor, since $\ln(e) = 1$.

A.E
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You could use another logarithm, say $\log_a$, but then you will have to differentiate $\log_ax$ which gives $\frac{1}{x\ln a}$; if $a = e$, this is just $\frac{1}{x}$. If you do the calculation with a base other than $e$, you still get the correct answer as there is a factor of $\frac{1}{\ln a}$ on both sides of the equation which cancels out.

Note, any two logarithms are related by a constant factor as $\log_ax = \frac{\ln x}{\ln a}$; in particular, when using logarithms to solve equations, you apply the logarithm to both sides. Then you will have, as above, a factor of $\frac{1}{\ln a}$ present on both sides which will cancel. As it cancels anyway, it is easier to choose $a = e$ so that the factor is simply $1$, effectively saving you a step in your working.

Michael Albanese
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