The power function
$x\mapsto x^m$ has the antiderivative $x\mapsto x^{m+1}/(m+1)$
for any rational number $m\neq -1$.
The function $x\mapsto x^{-1}=1/x$ is continuous on the positive real
axis and therefore has an antideriviative. The particular antiderivative which
takes on the value $0$ at the point $x=1$ is called the {\it natural
  logarithm} and is given by the formula

$$
\ln x=\int_1^x \dfrac {dt}t, \qquad x>0.
$$
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For all real numbers $x$ and $y$ and all rational numbers $r$ we have
$$
\ln(xy)=\ln x+\ln y,\quad
\ln(x/y)=\ln x -\ln y \quad \text{ og } \quad
\ln (x^r)=r\ln x.
$$

The natural logarithm is strictly increasing and we have the limits
$$
\lim_{x\to 0+}\ln x=-\infty \qquad \text{ og } \qquad
\lim_{x\to +\infty}\ln x=+\infty.
$$
This implies that $\ln$ has an inverse, which we denote by
$\exp$. The inverse turns out to be an exponential function $\exp
x=e^x$ with a number, $e$, as base. This number has the property that
$\ln e=1$. Any exponential function $x\mapsto a^x$ with base
$a>0$ can be written as
$a^x=e^{(\ln a)x}$.




\textbf{Example 10.8.1}\\
Simplify $ \ln(20) - 2\ln(2) + \ln(3)$.\\
\textbf{Solution:} By rules for logarithm we get
\[
    \ln(20) - 2 \ln(2) + \ln(3)
    = \ln(20) - \ln(2^2) + \ln(3)
    = \ln(20 \cdot 3) - \ln(4)
    = \ln \left( \frac{60}{4} \right) = \ln(15).
\]