Natural Logarithm Function

In many scientific fields, we need to solve equations where the unknown is an exponent, such as $e^x = a$. To do this, we use the inverse function of the exponential: the natural logarithm. While the exponential function grows extremely fast, the logarithm grows very slowly.

Definition and Link with Exponential

The exponential function is continuous and strictly increasing from $\mathbb{R}$ to $(0, +\infty)$. By the bijection theorem, for any strictly positive real number $x$, there exists a unique real number $y$ such that $e^y = x$. This unique solution $y$ is called the natural logarithm of $x$ and is denoted by $\ln(x)$.

Definition Natural Logarithm

The natural logarithm function, denoted by $\ln$, is the function defined on $(0, +\infty)$ which associates each real number $x > 0$ with the unique solution of the equation $e^y = x$ where $y$ is the unknown. We write:$$ y = \ln(x) $$

Proposition Reciprocity

For all $x > 0$ and any real $y$:

$e^y = x \iff y = \ln(x)$
$e^{\ln(x)} = x$
$\ln(e^y) = y$

Particular values:
$\ln(1) = 0$ (since $e^0 = 1$) and $\ln(e) = 1$ (since $e^1 = e$)

Proposition Symmetry

The graphs of the exponential and natural logarithm functions are symmetric with respect to the line $y = x$.

Algebraic Properties

Proposition Functional Relation

For all strictly positive real numbers $a$ and $b$:$$\ln(a \times b) = \ln(a) + \ln(b)$$

Proof

For all strictly positive real numbers $a$ and $b$:$$ \begin{aligned}e^{\ln(ab)} &= ab && \text{(by definition: } x = e^{\ln(x)}\text{)} \\ e^{\ln(ab)} &= e^{\ln(a)} \times e^{\ln(b)} && \text{(since } a = e^{\ln(a)} \text{ and } b = e^{\ln(b)}\text{)} \\ e^{\ln(ab)} &= e^{\ln(a) + \ln(b)} && \text{(property: } e^x \times e^y = e^{x+y}\text{)} \\ \ln(ab) &= \ln(a) + \ln(b)&& \text{(property: } e^x =e^y \iff x=y\text{)} \\ \end{aligned} $$

Proposition Other Algebraic Rules

For all $a > 0$, $b > 0$ and any integer $n$:

Inverse: $\ln\left(\frac{1}{b}\right) = -\ln(b)$
Quotient: $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$
Power: $\ln(a^n) = n\ln(a)$
Square Root: $\ln(\sqrt{a}) = \frac{1}{2}\ln(a)$

Proof

Inverse Rule: For $b > 0$: $$ \begin{aligned} e^{\ln(1/b)} &= \frac{1}{b} && \text{(by definition: } x = e^{\ln(x)}\text{)} \\ e^{\ln(1/b)} &= \frac{1}{e^{\ln(b)}} && \text{(since } b = e^{\ln(b)}\text{)} \\ e^{\ln(1/b)} &= e^{-\ln(b)} && \text{(property: } 1/e^x = e^{-x}\text{)} \\ \ln(1/b) &= -\ln(b) && \text{(property: } e^x = e^y \iff x=y\text{)} \end{aligned} $$
Quotient Rule: For $a > 0$ and $b > 0$: $$ \begin{aligned} e^{\ln(a/b)} &= \frac{a}{b} && \text{(by definition: } x = e^{\ln(x)}\text{)} \\ e^{\ln(a/b)} &= \frac{e^{\ln(a)}}{e^{\ln(b)}} && \text{(since } a = e^{\ln(a)} \text{ and } b = e^{\ln(b)}\text{)} \\ e^{\ln(a/b)} &= e^{\ln(a) - \ln(b)} && \text{(property: } e^x / e^y = e^{x-y}\text{)} \\ \ln(a/b) &= \ln(a) - \ln(b) && \text{(property: } e^x = e^y \iff x=y\text{)} \end{aligned} $$
Power Rule: For $a > 0$ and $n \in \mathbb{Z}$: $$ \begin{aligned} e^{\ln(a^n)} &= a^n && \text{(by definition: } x = e^{\ln(x)}\text{)} \\ e^{\ln(a^n)} &= (e^{\ln(a)})^n && \text{(since } a = e^{\ln(a)}\text{)} \\ e^{\ln(a^n)} &= e^{n\ln(a)} && \text{(property: } (e^x)^n = e^{nx}\text{)} \\ \ln(a^n) &= n\ln(a) && \text{(property: } e^x = e^y \iff x=y\text{)} \end{aligned} $$
Square Root Rule: For $a > 0$: $$ \begin{aligned} e^{\ln(\sqrt{a})} &= \sqrt{a} = a^{1/2} && \text{(by definition and power form)} \\ e^{\ln(\sqrt{a})} &= (e^{\ln(a)})^{1/2} && \text{(since } a = e^{\ln(a)}\text{)} \\ e^{\ln(\sqrt{a})} &= e^{\frac{1}{2}\ln(a)} && \text{(property: } (e^x)^n = e^{nx}\text{)} \\ \ln(\sqrt{a}) &= \frac{1}{2}\ln(a) && \text{(property: } e^x = e^y \iff x=y\text{)} \end{aligned} $$

Study of the Function

Proposition Derivative and Continuity

The function $\ln$ is differentiable and continuous on $(0, +\infty)$. For all $x > 0$:$$\ln'(x) = \frac{1}{x}$$Since $1/x > 0$ on its domain, the natural logarithm is strictly increasing on $]0, +\infty[$.

Proof

We assume (admit) that the function $\ln$ is differentiable on $(0, +\infty)$. We start from the identity:$$e^{\ln(x)} = x$$We differentiate both sides with respect to $x$:

On the right side: $(x)' = 1$.
On the left side, using the chain rule $(e^{u(x)})' = u'(x)e^{u(x)}$: $$\left(e^{\ln(x)}\right)' = \ln'(x) \times e^{\ln(x)} = \ln'(x) \times x$$

By equating the two results, we get:$$ \begin{aligned}\ln'(x) \times x &= 1 \\ \ln'(x) &= \frac{1}{x} && \text{(since } x > 0\text{)}\end{aligned} $$

Proposition Limits and Asymptote

The natural logarithm has the following end behaviors:

$\displaystyle\lim_{x \to +\infty} \ln(x) = +\infty$
$\displaystyle\lim_{x \to 0^+} \ln(x) = -\infty$

The line $x = 0$ (the y-axis) is a vertical asymptote to the curve of $\ln$.

Proposition Derivative of $\ln(u(x))$

Let $u$ be a differentiable and strictly positive function on an interval $I$. The function $f(x) = \ln(u(x))$ is differentiable on $I$ and:$$\textcolor{colorprop}{(\ln(u(x)))' = \frac{u'(x)}{u(x)}}$$

Comparative Growth

Proposition Comparative Growth

For any integer $n \ge 1$:

$\displaystyle\lim_{x \to +\infty} \dfrac{\ln(x)}{x} = 0$
$\displaystyle\lim_{x \to 0^+} x \ln(x) = 0$
$\displaystyle\lim_{x \to +\infty} \dfrac{\ln(x)}{x^n} = 0$
$\displaystyle\lim_{x \to 0^+} x^n \ln(x) = 0$

Proof

For $x > 0$, $\dfrac{\ln(x)}{x} = \dfrac{\ln(x)}{e^{\ln(x)}}$. Let $X = \ln(x)$. As $x \to +\infty$, $X \to +\infty$. By the composition theorem and the known limit $\displaystyle\lim_{X \to +\infty} \dfrac{X}{e^X} = 0$, we conclude that $\displaystyle\lim_{x \to +\infty} \dfrac{\ln(x)}{x} = 0$.
For $x > 0$, $x \ln(x) = e^{\ln(x)} \times \ln(x)$. Let $X = \ln(x)$. As $x \to 0^+$, $X \to -\infty$. By the composition theorem and the known limit $\displaystyle\lim_{X \to -\infty} X e^X = 0$, we conclude that $\displaystyle\lim_{x \to 0^+} x \ln(x) = 0$.
For $n > 1$, we can write $\dfrac{\ln(x)}{x^n} = \dfrac{1}{x^{n-1}} \times \dfrac{\ln(x)}{x}$. Since $\displaystyle\lim_{x \to +\infty} \dfrac{1}{x^{n-1}} = 0$ and $\displaystyle\lim_{x \to +\infty} \dfrac{\ln(x)}{x} = 0$, the product of the limits gives $0$.
For $n > 1$, we can write $x^n \ln(x) = x^{n-1} \times x \ln(x)$. Since $\displaystyle\lim_{x \to 0^+} x^{n-1} = 0$ and $\displaystyle\lim_{x \to 0^+} x \ln(x) = 0$, the product of the limits gives $0$.