A calculus of the absurd

Chapter 9 Exponentials and logarithms

9.1 Exponentials

Hopefully you’re vaguely aware that \(a^b\) means "a multiplied by itself b times" (for \(b \in \mathbb {N}\))6161 If you’re not sure about this notation, review the section on "Sets and Numbers". From this definition, there are a bunch of useful facts we can derive.

\begin{equation} a^{b + c} = a^{b}a^{c} \end{equation}

A somewhat non-rigorous argument for this being true is as follows: \(a^b = a * a * a * ... * a\) (a times itself b times). When we multiply \(a^b\) by \(a^c\), which is equal to \(a^c = a * a * a * ... * a\) (a times itself c times), we are then multiplying \(a\) times itself \(b\) times by \(a\) times itself \(c\) times. Overall, therefore we are multiplying \(a\) by itself \(b+c\) times.

\begin{equation} \left (a^{b}\right )^{c} = a ^ {bc} \end{equation}

To see that this is true, first note that we start by multiplying \(a\) by itself \(b\) times (\(a*a*a*...*a\)). We then raise this to the power of \(c\), so \((a*a*a*...*a) ^ c\), which means we have \((a*a*a*...*a) * (a*a*a*...*a) * ... * (a*a*a*...*a)\). In total, there are \(c * b\) lots of \(a\) (every bracket is \(b\) lots of \(a\), and there are \(c\) of the brackets, so overall there are \(c * b\) lots of \(a\)).