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As discussed in this question, there are many different approaches to defining the natural logarithm function. In particular, since the exponential function $ \exp(x) := \sum_{k=0}^{\infty}\frac{x^k}{k!} $ is strictly increasing, its inverse exists and by definition $ \ln(x) := \exp^{-1}(x). $ On the other hand, then natural logarithm can also be defined through $ \ln(x) := \int_1^{x}\frac{1}{t}dt. $

What is not at all obvious to me is how these two definitions are equivalent. So, my first question is, how is it that these two definitions are equivalent? A related question is, if one wanted to modify either of these definitions to account for a base other than $e$, how would one proceed? Note that a reference that discusses these topics is perfectly acceptable answer.

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    Closely related to http://math.stackexchange.com/a/31392/589 if not a duplicate.2012-03-22

2 Answers 2

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I guess this is worth writing down. By your definition we have $\text{exp}(\log(x)) = x$. Differentiating gives \text{exp}(\log(x)) \log'(x) = x \log'(x) = 1

hence \log'(x) = \frac{1}{x}. The desired result then follows by the fundamental theorem of calculus and the observation that $\log(1) = 0$.

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In your link you may find a variation of : $\log(x) = \lim_{n\to \infty} l_n\ \ \text{with }\ l_n=n\cdot\left(x^{\frac 1n}-1\right)$

Let's revert the equation at the right : $\displaystyle 1+\frac {l_n}n=x^{\frac 1n}$ so that $x=\left( 1+\frac {l_n}n\right)^n\ \text{with }\lim_{n\to \infty} \left( 1+\frac {l_n}n\right)^n=e^{\log(x)}$

getting I hope a clear symmetry.

The last term is too $\displaystyle \lim_{n\to \infty} \left( 1+\frac {\log(x)}n\right)^n$ : a classical definition of exponential which becomes at the limit your first formula : $e^{\log(x)} = \sum_{k=0}^{\infty}\frac{(\log(x))^k}{k!}$

Concerning $l_n=n\cdot\left(x^{\frac 1n}-1\right)$ and the classical $\ln(x) = \int_1^{x}\frac{1}{t}dt$ let's observe that : $\log(x)=\int_1^{x}\frac{1}{t}dt= \lim_{n\to \infty} \int_1^x\frac 1{t^{1-\frac1n}}dt=\lim_{n\to \infty}\left[\frac {t^{\frac 1n}}{\frac 1n}\right]_1^x=\lim_{n\to \infty} n\cdot\left(x^{\frac 1n}-1\right)$

To get the classical Taylor series for $\log$ you may use the last limit with $x$ replaced by $1+t$.

For the second question see Robert's comment.

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    @Peter: Thanks for that Peter. Cheers,2012-03-22