## Exponent Laws

There are several laws we can use to make working with exponential numbers easier. Some of these laws might be already be familiar to you, but we list all the laws here for easy reference:

\({a}^{m}\times {a}^{n}={a}^{m+n}\)

\(\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}\)

\({(ab)}^{n}={a}^{n}{b}^{n}\)

\({(\dfrac{a}{b})}^{n}=\dfrac{{a}^{n}}{{b}^{n}}\)

\({({a}^{m})}^{n}={a}^{mn}\)

where \(a>0\), \(b>0\) and \(m,n\in \mathbb{R}\)

The following two videos explain the exponent laws.

### Part 1:

### Part 2:

## Example

### Question

Simplify:

\({2}^{3x}\times {2}^{4x}\)

\(\dfrac{4x^{3}}{2x^{5}}\)

\(\dfrac{12{p}^{2}{t}^{5}}{3p{t}^{3}}\)

\({(3x)}^{2}\)

\({({3}^{4}{5}^{2})}^{3}\)

\(6p^{0} \times (7p)^{0}\)

\({(\dfrac{2xp}{6x^2})}^{3}\)

\({({2}^{-2})}^{2x+1}\)

\({2}^{3x}\times {2}^{4x}={2}^{3x+4x}={2}^{7x}\)

\(\dfrac{4x^{3}}{2x^{5}}=2x^{3-5}=2x^{-2}=\dfrac{2}{x^{2}}\)

\(\dfrac{12{p}^{2}{t}^{5}}{3p{t}^{3}}=4{p}^{(2-1)}{t}^{(5-3)}=4p{t}^{2}\)

\({(3x)}^{2}={3}^{2}{x}^{2}=9{x}^{2}\)

\({({3}^{4}\times {5}^{2})}^{3}={3}^{(4\times 3)}\times {5}^{(2\times 3)}={3}^{12}\times {5}^{6}\)

\(6p^{0} \times (7p)^{0} = 6(1) \times 1 = 6\)

\({(\dfrac{2xp}{6x^2})}^{3}=(\dfrac{p}{3x})^{3}=\dfrac{p^{3}}{27x^{3}}\)

\({({2}^{-2})}^{2x+1}=2^{-2(2x+1)}=2^{-4x-2}\)

## Example

### Question

Simplify: \(\dfrac{{2}^{2n}\times {4}^{n}\times 2}{{16}^{n}}\)

### Change the bases to prime numbers

At first glance it appears that we cannot simplify this expression. However, if we reduce the bases to prime bases, then we can apply the exponent laws.

\[\dfrac{{2}^{2n} \times {4}^{n} \times 2}{{16}^{n}} = \dfrac{{2}^{2n} \times {({2}^{2})}^{n} \times {2}^{1}}{{({2}^{4})}^{n}}\]

### Simplify the exponents

\begin{align*} & =\dfrac{{2}^{2n} \times {2}^{2n} \times {2}^{1}}{{2}^{4n}} \\ & = \dfrac{{2}^{2n + 2n + 1}}{{2}^{4n}} \\ & = \dfrac{{2}^{4n + 1}}{{2}^{4n}} \\ & ={2}^{4n + 1 – (4n)} \\ & = 2 \end{align*}

#### Tip:

When you have a fraction that is one term over one term, use the method of Finding Prime Bases – in other words use prime factorisation on the bases.

## Example

### Question

Simplify:

\[\dfrac{{5}^{2x – 1} \cdot {9}^{x – 2}}{{15}^{2x – 3}}\]

### Change the bases to prime numbers

\begin{align*} \dfrac{{5}^{2x – 1} \cdot {9}^{x – 2}}{{15}^{2x – 3}} & = \dfrac{{5}^{2x – 1} \cdot {({3}^{2})}^{x – 2}}{{(5\times 3)}^{2x – 3}} \\ & = \dfrac{{5}^{2x – 1} \cdot {3}^{2x – 4}}{{5}^{2x – 3} \cdot {3}^{2x – 3}} \end{align*}

### Subtract the exponents (same base)

\begin{align*} & = {5}^{(2x – 1) – (2x – 3)} \times {3}^{(2x – 4) – (2x – 3)} \\ & = {5}^{2x – 1 – 2x + 3} \times {3}^{2x – 4 – 2x + 3} \\ & = {5}^{2} \times {3}^{-1} \end{align*}

### Write the answer as a fraction

\begin{align*} &= \dfrac{{5}^{2}}{3} \\ & = \dfrac{25}{3} \end{align*}

#### Fact:

When working with exponents, all the laws of operation for algebra apply.

## Example

### Question

Simplify:

\[\dfrac{{2}^{t} – {2}^{t – 2}}{3\cdot {2}^{t} – {2}^{t}}\]

### Simplify to a form that can be factorised

For each of the exponent laws we can “undo” the law – in other words we can work backwards. For this expression we can reverse the multiplication law to write \(2^{t – 2}\) as \(2^{t} \cdot 2^{-2}\).

\[\cfrac{{2}^{t} – {2}^{t – 2}}{3\cdot {2}^{t} – {2}^{t}} = \cfrac{{2}^{t} – ({2}^{t}\cdot {2}^{-2})}{3\cdot {2}^{t}-{2}^{t}}\]

### Take out a common factor

\[= \cfrac{{2}^{t}(1 – {2}^{-2})}{{2}^{t}(3 – 1)}\]

### Cancel the common factor and simplify

\begin{align*} & = \cfrac{1-2^{-2}}{3-1} \\ & = \cfrac{1 – \cfrac{1}{4}}{2}\\ & = \cfrac{3}{4} \times \cfrac{1}{2} \\ & = \cfrac{3}{8} \end{align*}

#### Tip:

When you have a fraction that has more than one term in the numerator or denominator, change to prime bases if necessary and then factorise.

## Example

### Question

Simplify:

\[\dfrac{{9}^{x} – 1}{{3}^{x} + 1}\]

### Change the bases to prime numbers

\begin{align*} \dfrac{{9}^{x} – 1}{{3}^{x} + 1} & = \dfrac{{({3}^{2})}^{x} – 1}{{3}^{x} + 1} \\ & = \dfrac{{3}^{2x} – 1}{{3}^{x} + 1} \qquad \text{Recognise that } 3^{2x} = (3^{x})^{2} \end{align*}

### Factorise using the difference of squares

\[= \dfrac{({3}^{x} – 1)({3}^{x} + 1)}{{3}^{x} + 1}\]

### Cancel the common factor and simplify

\[= {3}^{x}-1\]