Deriving properties of the logarithm from its integral representation

Question

Suppose we define:

$$\ln(x) = \int_{a}^{x} \left[ \frac{1}{r} dr\right]$$ Such that

$$ \ln(1) = 0, \ln(e) = 1$$

How does one derive all the properties of the logarithm from the properties of the function $$ \frac{1}{x}$$

For example:

If we let $x,y$ be functions of t. Then it follows that

$$ \frac{1}{x} \frac{dx}{dt} + \frac{1}{y} \frac{dy}{dt} = \frac{1}{xy} \left( y \frac{dx}{dt} + x \frac{dy}{dt} \right) $$

Thus

$$ \ln(x(t)) + \ln(y(t)) = \ln(x(t)y(t)) $$

Arriving at the above statement was artificial. I first assumed the latter property true and then differentiated to find what initial algebraic identitiy had to exist. Is there a more natural motivation for this?

Answer 1

Another way to prove that product formula:

$$\log(ax)=\int_{1}^{ax} \frac{dt}{t} =\int_{1}^x \frac{dt}{t} + \int_{x}^{ax} \frac{dt}{t} = \log(x) + \int_{x}^{ax} \frac{dt}{t}$$

Substituting $t=xu$, we get $$\int_{x}^{ax} \frac{dt}{t} = \int_{1}^{a}\frac{du}{u}=\log(a)$$

By induction from here, you can show $\log(x^n)=n\log(x)$.

If you define $e=\lim_{n\to\infty} \left(1+\frac1n\right)^n$, then:

$$\log(e) =\lim_{n\to\infty} \frac{\log\left(1+\frac{1}{n}\right)}{1/n}$$

The right hand is (essentially) the definition of the deivative of $\log(x)$ at $x=1$, so the limit is $\frac{1}{1}$ by the Fundamental Theorem of Calculus.

Answer 2

Define a function $f$ as

$$f(x) = \int_1^{x} \frac{1}{t} \,dt$$

for $x>0$. It is easy to see show the following properties.

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Property (1): The function $f$ is increasing.

To show this, assume $x_2>x_1>0$, and form the difference $f(x_2)-f(x_1)=\int_{x_1}^{x_2} \frac{1}{t}\,dt$. Note that this difference is strictly positive since the integrand is also strictly positive. Thus, $f(x_2)>f(x_1)$ whenever $x_2>x_1$, which implies $f$ is strictly increasing.

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Property (2): The function $f$ is differentiable with $f'(x)=\frac{1}{x}$.

This follows immediately from the fundamental theorem of calculus. Note that this implies automatically that $f$ is continuous for $x>0$.

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Property (3): The limit $\lim_{x \to \infty} f(x)=+\infty$

We can easily verify that $\lim_{x \to \infty} f(x)=+\infty$ since for $n>1$

$$\begin{align} \int_1^{2^n} \frac{1}{t} dt&=\int_{1}^{2} \frac{1}{t} \,dt+\int_{2}^{4} \frac{1}{t} dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{t} \,dt\\\\ &\ge \int_{1}^{2} \frac{1}{2} \,dt+\int_{2}^{4} \frac{1}{4} \,dt+\cdots +\int_{2^{n-1}}^{2^{n}} \frac{1}{2^n} \,dt\\\\ &=\frac12 (2-1)+\frac14 (4-2) + \cdots + \frac{1}{2^{n}}(2^n-2^{n-1})\\\\ &=\frac{n}{2} \end{align}$$

which tends to infinity as $n \to \infty$.

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Property (4): The limit $\lim_{x \to 0} f(x)=-\infty$

Examine the integral

$$f(\frac{1}{n})=-\int_{\frac{1}{n}}^1 \frac{1}{t}\,dt$$ and substitute $u=\frac{1}{t}$. Then, we have $$-\int_{\frac{1}{n}}^1 \frac{1}{t}\,dt=-\int_1^n \frac{1}{u}\,du$$

But from property (3), this integral tends to $+\infty$ and thus $f(\frac1{n})$ tends to $-\infty$ as $n \to 0$.

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Property (5): The function $f$ satisfies $f(x^n) = nf(x)$.

We know by definition that $$f(x^n) = \int_1^{x^n} \frac{1}{t} \,dt$$

Then, by substituting $t=u^n$, with $dt=nu^{n-1}du$, and the limits going from $1$ to $x$ reveals $$\begin{align} \log (x^n) &=\int_1^{x} \frac{nu^{n-1}}{u^n} \,du\\\\ &=\int_1^{x} n\frac{1}u\,du\\\\ &=nf (x) \end{align}$$

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The other well-known properties for the log of a product and quotient follow from Property (5). For example, in THIS ANSWER, I show that $f(xy)=f(x)+f(y)$.