# Powers, exponents and logarithms of natural numbers

## Powers

### Recap

Previously we defined addition and multiplication in terms of successive use of the sucessor function. That is, the definition of addition was:

$$\forall a \in \mathbb{N} (a+0=a)$$

$$\forall a b \in \mathbb{N} (a+s(b)=s(a+b))$$

And similarly for multiplication:

$$\forall a \in \mathbb{N} (a.0=0)$$

$$\forall a b \in \mathbb{N} (a.s(b)=a.b+a)$$

Additional functions could also be defined, following the same pattern:

$$\forall a \in \mathbb{N} (a\oplus _n 0=c)$$

$$\forall a b \in \mathbb{N} (a\oplus _{n} s(b)=(a\oplus_{n} b)\oplus_{n-1}a)$$

### Powers

Powers can also be defined:

$$\forall a \in \mathbb{N} a^0=1$$

$$\forall a b \in \mathbb{N} a^{s(b)}=a^b.a$$

### Example

So $$2^2$$ can be calculated like:

$$2^2=2^{s(1)}$$

$$2^{s(1)}=2.2^1$$

$$2.2^1=2.2.2^0$$

$$2.2.2^0=2.2.1$$

$$2.2.1=4$$

Unlike addition and multiplication, exponention is not commutative. That is

$$a^b\ne b^a$$

### Exponential rules

$$a^ba^c=a^{b+c}$$

$$(a^b)^c=a^{bc}$$

$$(ab)^c=a^cb^c$$

### Binomial expansion

How can we expand

$$(a+b)^n, n\in \mathbb{N}$$

We know that:

$$(a+b)^n=(a+b)(a+b)^{n-1}$$

$$(a+b)^n=a(a+b)^{n-1}+b(a+b)^{n-1}$$

Each time this is done, the terms split, and each terms is multiplied by either $$a$$ or $$b$$. That means at the end there are $$n$$ total multiplications.

This can be shown as:

$$(a+b)^n=\sum_{i=1}^n a^i b^{n-i} c_i$$

So we want to identify $$c_i$$.

Each term can be shown as a series of $$n$$ $$a$$s and $$b$$s. For example:

• $$aaba$$

• $$baaa$$

For any of these, there are $$n!$$ ways or arranging the sequence, but this includes duplicates. If we were given $$n$$ unique terms to multiply there would indeed by $$n!$$ different ways this could have arisen, but we can swap $$a$$s and $$b$$s, as they were only generated once. So let’s count duplicates.

There are duplicates in the $$a$$s. If there are $$i$$ $$a$$s, then there are $$i!$$ ways of rearranging this. Similarly, if there are $$n-i$$ $$b$$s, then there are $$(n-i)!$$ ways or arranging this.

As a result the number of actual observed instances, $$c_i$$, is:

$$c_i=\dfrac{n!}{i!(n-i)!}$$

And so:

$$(a+b)^n=\sum^n_{i=0} a^i b^{n-i} \dfrac{n!}{i!(n-i)!}$$

We can also write this last term as:

$$\begin{pmatrix}n\\i\end{pmatrix}$$

### Difference of two squares

$$(a+b)(a-b)=a^2-ab+ab-b^2$$

$$(a+b)(a-b)=a^2-b^2$$

## Logarithms

### Definition

If:

$$c=a^b$$

Then

$$log_ac=b$$

Product rule:

$$a=c^{log_ca }$$

$$b=c^{log_cb }$$

So:

$$ab=c^{log_cab }$$

But also:

$$ab=c^{log_ca }c^{log_cb }$$

$$ab=c^{log_ca + log_cb }$$

So:

$$log_ca+log_cb=log_cab$$

### Power rule

$$a=b^{log_ba}$$

So:

$$a^c=b^{log_ba^c}$$

And separately:

$$a^c=(b^{log_ba})^c$$

$$a^c=(b^{clog_ba})$$

So:

$$clog_ba=log_ba^c$$