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-# Inverse Functions and the Logarithm
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+# Inverse Functions and the Logarithm
+
+## Inverse Function
+
+If $f$ is a function, then the function $g$ is the inverse function of $f$ if
+
+$$g(f(x))=x$$
+
+for all $x$ in which $f(x)$ can be calculated.
+
+### Details
+
+The inverse of a function $f$ is denoted by $f^{-1}$, i.e.
+
+$$f^{-1}(f(x))=x$$
+
+### Examples
+
+:::info Example
+
+If $f(x) = x^2$ for $x<0$ then the function $g$, defined as $g(y)=\sqrt{y}$ for $y>0$, is not the inverse of $f$ since $g(f(x))=\sqrt{x^2} = |x| = -x$ for $x<0$.
+
+:::
+
+## When the Inverse Exists: The Domain Question
+
+Inverses do not always exist.
+For an inverse of $f$ to exist, $f$ must be one-to-one, i.e. for each $x$, $f(x)$ must be unique.
+
+![Fig. 13](../media/10_2_When_the_inverse_exists.png)
+
+Figure: The function $f(x)=x^2$ does not have an inverse since $f(x)=1$
+
+has two possible solutions $-1$ and $1$.
+
+### Examples
+
+:::info Example
+
+$f(x)=x^2$ does not have an inverse since $f(x)=1$ has two possible solutions -1 and 1.
+
+:::
+
+:::note Note
+
+Note that iff $f$ is a function, then the function $g$ is the inverse function of $f$, if $g(f(x)) = x$ for all calculated values of $x$ in $f(x)$.
+
+The inverse function of $f$ is denoted by $f^{-1}$, i.e. $f^{-1}(f(x)) = x$.
+
+:::
+
+:::info Example
+
+What is the inverse function, $f^{-1}$, of $f$ if $f(x) = 5 + 4x$.
+The simplest approach is to write $y=f(x)$ and solve for $x$.
+
+With
+
+$$f(x) = 5 + 4x$$
+
+we write
+
+$$y = 5 + 4x$$
+
+which we can now rewrite as
+
+$$y - 5 = 4x$$
+
+and this implies
+
+$$\displaystyle\frac{y-5}{4} = x$$
+
+And there we have it, very simple:
+
+$$f^{-1}(f(x)) = \displaystyle\frac{y - 5}{4}$$
+
+:::
+
+## The Base 10 Logarithm
+
+When $x$ is a positive real number in $x=10^y$, $y$ is referred to as the base 10 logarithm of x and is written as:
+
+$$y=\log_{10}(x)$$
+
+or
+
+$$y=\log(x)$$
+
+### Details
+
+If $\log (x) = a$ and $\log (y)=b$, then $x = 10^a$ and $y = 10^b$, and
+
+$$x \cdot y = 10^a \cdot 10^b = 10^{a+b}$$
+
+so that
+
+$$\log(xy) = a+b$$
+
+### Examples
+
+:::info Example
+
+$$
+\begin{aligned}
+  \log(100) &= 2 \\
+  \log(1000) &= 3
+\end{aligned}
+$$
+
+:::
+
+:::info Example
+
+If
+
+$$\log(2) \approx 0.3$$
+
+then
+
+$$10^y=2$$
+
+:::
+
+:::note Note
+
+$$2^{10}=1024 \approx 1000 = 10^3$$
+
+therefore
+
+$$2 \approx 10^{3/10}$$
+
+so
+
+$$\log (2) \approx 0.3$$
+
+:::
+
+## The Natural Logarithm
+
+A logarithm with $e$ as a base is referred to as the natural logarithm and is denoted as $\ln$:
+
+$$y=\ln(x)$$
+
+if
+
+$$x=e^y=\exp(y)$$
+
+Note that $\ln$ is the inverse of $\exp$.
+
+![Fig. 14](../media/10_4_The_natural_logarithm.png)
+
+Figure: The curve depicts the function $y=\ln(x)$ and shows that $\ln$ is the inverse of $\exp$.
+Note that $\ln(1)=0$ and when $y=0$ then $e^0=1$.
+
+## Properties of Logarithm(s)
+
+Logarithms transform multiplicative models into additive models, i.e.
+
+$$\ln(a\cdot b) = \ln a + \ln b$$
+
+### Details
+
+This implies that any statistical model, which is multiplicative becomes additive on a log scale, e.g.
+
+$$y = a \cdot w^b \cdot x^c$$
+
+$$\ln y = (\ln a) + \ln (w^b) + \ln (x^c)$$
+
+Next, note that
+
+$$
+\begin{aligned}
+  \ln (x^2) &= \ln (x \cdot x) \\
+  &= \ln x + \ln x \\
+  &= 2 \cdot \ln x
+\end{aligned}
+$$
+
+and similarly $\ln (x^n) = n \cdot \ln x$ for any integer $n$.
+
+In general $\ln (x^c) = c \cdot \ln x$ for any real number c (for $x>0$).
+
+Thus the multiplicative model (from above)
+
+$$y=a \cdot w^b \cdot x^c$$
+
+becomes
+
+$$y= (\ln a) + b \cdot \ln w + c \cdot \ln x$$
+
+which is a linear model with parameters $(\ln a)$, $b$ and $c$.
+
+In addition, the log-transform is often variance-stabilizing.
+
+## The Exponential Function and the Logarithm
+
+The exponential function and the logarithms are inverses of each other
+
+$$x = e^y \leftrightarrow y = \ln{x}$$
+
+### Details
+
+:::note Note
+
+Note the properties:
+
+$$\ln (x \cdot y) = \ln (x) + \ln (y)$$
+
+and
+
+$$e^a \cdot e^b = e^{a+b}$$
+
+:::
+
+### Examples
+
+:::info Example
+
+Solve the equation
+
+$$10e^{1/3x} + 3 = 24$$
+
+for $x$.
+
+First, get the $3$ out of the way:
+
+$$10e^{1/3x} = 21$$
+
+Then the $10$:
+
+$$e^{1/3x} = 2.1$$
+
+Next, we can take the natural log of 2.1.
+Since $\ln$ is an inverse function of $e$ this would result in
+
+$$\displaystyle\frac{1}{3}x = \ln(2.1)$$
+
+This yields
+
+$$x = \ln(2.1) \cdot 3$$
+
+which is
+
+$$\approx 2.23$$
+
+:::