I am trying to put together a more rigorous view of logarithms and exponentials than what is typically taught in a first-year calculus course. I know that we can define

$$\ln(x) = \int_{1}^x \frac{1}{t}dt,$$

and we can use this to show that this function satisfies the usual properties for logarithms, but does that mean it IS a logarithm?

Other approaches I have seen show that $\ln(x)$ is invertible, and define its inverse to be $e^x$, but this seems to raise the same question. Since we can prove that

$$e= \lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^n,$$

using the integral definition of $\ln(x)$, it seems unwise to take this as a definition for $e$ and use that to argue that $\ln(x)$ is a logarithm.

Does anyone know of compatible definitions for $e$ and $\ln(x)$ that allow us to prove that $\ln(x)$ is a logarithm and that $e^x$ is its inverse without becoming circular? For example, can one use the integral definition to find a change of base formula for logs? Can you use this definition to show that $e^{\ln(x)}= x$?

  • $\begingroup$ What exactly would convince you that something "IS a logarithm", if not that it satisfies the usual properties for logarithms? $\endgroup$ – TonyK Sep 15 '16 at 19:49
  • $\begingroup$ I think I was always told that they are the inverses of exponential functions, hence, they satisfy these properties. However, with the integral definition, we usually pick $e$ to be the number that makes the integral equal 1. Then it is easy to prove that $\ln(e^x)=x$, but I don't know how to do the other composition. Maybe I don't need to because exponentials are bijective? $\endgroup$ – Taylor Sep 15 '16 at 20:12
  • $\begingroup$ You may want to prove that, given $a>0$, there exists one and only one function $g_a\colon\mathbb{R}\to\mathbb{R}_{>0}$ such that $g_a(x+y)=g_a(x)g_a(y)$ for all $x,y\in\mathbb{R}$, and $g_a(1)=a$. $\endgroup$ – egreg Sep 15 '16 at 20:16
  • $\begingroup$ Should I prove that or prove the analagous rule for logarithms? $\endgroup$ – Taylor Sep 15 '16 at 20:19
  • $\begingroup$ This question requires an answer too long to be expected here. I suggest that you look at the following two books. Calculus: Volume 1 by Tom Apostol. Elementary Real and Complex Analysis by Georgi Shilov. $\endgroup$ – poweierstrass Sep 15 '16 at 20:21

Maybe it helps to change notation. Define $L(x) = \displaystyle \int_1^x \frac 1t \, dt$. It is fairly routine to show that $L$ is differentiable, increasing, and maps $(0,\infty)$ onto $(-\infty,\infty)$. Thus $L$ has a differentiable inverse $L^{-1}$ that is increasing and maps $(-\infty,\infty)$ onto $(0,\infty)$. Define $E(x) = L^{-1}(x)$.

It follows that $E(L(x)) = x$ for all $x > 0$ and $L(E(x)) = x$ for all $x \in \mathbb R$.

Using the inverse function theorem you find that $E'(x) = \dfrac{1}{L'(E(x))} = E(x)$.

The change-of-variable theorem implies that $L(x^n) = nL(x)$ for all $x > 0$ and natural numbers $n$. Since $$ \frac{n}{n+x} \frac xn \le \int_1^{1+x/n} \frac 1t \, dt \le \frac xn$$ for all such $x$ and $n$ you get $$\frac{nx}{n+x} \le nL(1+x/n) \le x$$ so that $$\lim_{n \to \infty} L \left( \left(1+\frac xn\right)^n \right) = x.$$ Since $E$ and $L$ are both continuous you conclude that $(1+x/n)^n$ converges as $n \to \infty$ and moreover $$E(x) = \lim_{n \to \infty} \left( 1 + \frac xn \right)^n.$$

The computation needs to be adjusted slightly if $x < 0$ but the limit remains valid for all $x$.

  • $\begingroup$ The goal is to eventually differentiate $E(x)$ and $L(x)$ and find formula. The Fundamental Theorem of Calculus helps me with $L$, but how do I find a formula for $E'$? Moreover, once I have done this, how do I justify using these functions to find the derivative of arbitrary exponentials $B(x)$? $\endgroup$ – Taylor Sep 15 '16 at 20:23
  • $\begingroup$ @Taylor as stated in the answer use the inverse function theorem to evaluate $E'$. $\endgroup$ – Umberto P. Sep 15 '16 at 20:25
  • $\begingroup$ General exponential functions are defined by $B(x) = b^x = E[x L(b)]$. $\endgroup$ – Umberto P. Sep 15 '16 at 20:25
  • $\begingroup$ Hmm. Okay. This seems like it will work. $\endgroup$ – Taylor Sep 15 '16 at 20:27

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