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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    $\begingroup$ Take the difference of the two functions, differentiate, get $0$, note the two functions coincide at $x = 1$. $\endgroup$ – t.b. Mar 21 '12 at 20:52
  • $\begingroup$ It's obvious 8-). (For a more serious note: What t.b. said) $\endgroup$ – Jonas Teuwen Mar 21 '12 at 20:56
  • $\begingroup$ @ItsNotObvious Great username! $\endgroup$ – Jeff Mar 21 '12 at 21:06
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    $\begingroup$ As for other bases, note that $\log_a(x)$ is the inverse function of $$a^x = \exp(x \ln(a)) = \sum_{k=0}^\infty \frac{(\ln a)^k x^k}{k!}$$ while in the other approach $$ \log_a(x) = \frac{\ln(x)}{\ln(a)} = \frac{ \int_1^x \frac{1}{t}\ dt}{\int_1^a \frac{1}{t} dt}$$ $\endgroup$ – Robert Israel Mar 21 '12 at 21:17
  • $\begingroup$ Closely related to math.stackexchange.com/a/31392/589 if not a duplicate. $\endgroup$ – lhf Mar 22 '12 at 1:20

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 $\log(1) = 0$.


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.

  • $\begingroup$ I really liked the first insight! Good one. $\endgroup$ – Pedro Tamaroff Mar 22 '12 at 3:41
  • $\begingroup$ @Peter: Thanks for that Peter. Cheers, $\endgroup$ – Raymond Manzoni Mar 22 '12 at 6:52

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