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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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ItsNotObvious
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    Take the difference of the two functions, differentiate, get $0$, note the two functions coincide at $x = 1$. – t.b. Mar 21 '12 at 20:52
  • It's obvious 8-). (For a more serious note: What t.b. said) –  Mar 21 '12 at 20:56
  • @ItsNotObvious Great username! – Jeff Mar 21 '12 at 21:06
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    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 $$ \log_a(x) = \frac{\ln(x)}{\ln(a)} = \frac{ \int_1^x \frac{1}{t}\ dt}{\int_1^a \frac{1}{t} dt}$$ – Robert Israel Mar 21 '12 at 21:17
  • Closely related to http://math.stackexchange.com/a/31392/589 if not a duplicate. – lhf Mar 22 '12 at 01:20

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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 $\log(1) = 0$.

Qiaochu Yuan
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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.

Raymond Manzoni
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See Ricardo, H. (2022). The Equivalence of Definitions of the Natural Logarithm Function. The College Mathematics Journal, 53(3), 190–196. https://doi.org/10.1080/07468342.2022.2039553.

PolyaPal
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  • Links to external resources are encouraged, but please add context around the link so your fellow users will have some idea what it is and why it’s there. Always quote the most relevant part of an important link, in case the external resource is unreachable or goes permanently offline. – From [answer]. – Martin R Jun 25 '24 at 14:51