Equivalence of Logarithm Definitions 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.
 A: 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 $\log(1) = 0$. 
A: 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.
