# A calculus of the absurd

### Chapter 8 Exponentials and logarithms

#### 8.1 Exponentials

Hopefully you’re vaguely aware that $$a^b$$ means "a multiplied by itself b times" (for $$b \in \mathbb {N}$$)5555 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 $$a$$).