1. Indices and Surds

iGCSE (2021 Edition)

Lesson

When multiplying a number by itself repeatedly, we are able to use index notation to write the expression more simply. Here we are going to look at a rule that allows us simplify products that involve the multiplication of index terms.

Consider the expression $a^5\times a^3$`a`5×`a`3. Notice that the terms share like bases.

Let's think about what this would look like if we expanded the expression:

We can see that there are eight $a$`a`s being multiplied together, and notice that $8$8 is the sum of the powers in the original expression.

So, in our example above,

$a^5\times a^3$a5×a3 |
$=$= | $a^{5+3}$a5+3 |

$=$= | $a^8$a8 |

Let's look at a specific example. Say we wanted to find the value of $4^2\times4^3$42×43. By evaluating each product separately we would have

$4^2\times4^3$42×43 | $=$= | $16\times64$16×64 |

$=$= | $1024$1024 |

Alternatively, by first expanding the terms in the original expression we can arrive at a simplified version of the expression on our way to the final value.

$4^2\times4^3$42×43 | $=$= | $\left(4\times4\right)\times\left(4\times4\times4\right)$(4×4)×(4×4×4) |

$=$= | $4^5$45 | |

$=$= | $1024$1024 |

Notice in the second line we have identified that $4^2\times4^3=4^5$42×43=45.

We can avoid having to write each expression in expanded form by using the multiplication law.

The multiplication law

For any base number $a$`a`, and any numbers $m$`m` and $n$`n` as powers,

$a^m\times a^n=a^{m+n}$`a``m`×`a``n`=`a``m`+`n`

That is, when multiplying terms with a common base:

- Keep the same base
- Find the sum of the powers

When multiplying terms with like bases, we **add** the indices (or powers).

The multiplication law **only** works for terms with the **same bases**.

Consider the expression $7^2\times3^4$72×34.

$7$7 and $3$3 are not the same base terms, so we cannot simplify this expression any further.

But we can simplify the following expression: $7^2\times3^2$72×32 in another way.

Notice that both terms have the same base, so we can multiply their bases, and keep the same index..

$7^2\times3^2$72×32 | $=$= | $7\times7\times3\times3$7×7×3×3 |

$=$= | $\left(7\times3\right)\times\left(7\times3\right)$(7×3)×(7×3) | |

$=$= | $21\times21$21×21 | |

$=$= | $21^2$212 |

Multiplying terms with the same index

For any base numbers $a$`a` and $b$`b` that both have $n$`n` as a power,

$a^n\times b^n=\left(a\times b\right)^n$`a``n`×`b``n`=(`a`×`b`)`n`

That is, when multiplying terms with the same index:

- Find the product of the bases
- Keep the same power

Write $3^4\times3^{10}$34×310 in simplest index form.

Using index laws, evaluate $3^4\times3^3$34×33.

The method to divide power terms is similar to the multiplication law, however in this case we subtract the powers from one another, rather than add them. Let's look at an expanded example to see why this is the case.

If we wanted to simplify the expression $a^5\div a^2$`a`5÷`a`2, we could write it as:

We can see that there are five $a$`a`s being divided by two $a$`a`s to give a result of three $a$`a`s, and notice that $3$3 is the difference of the powers in the original expression.

So, in our example above,

$a^5\div a^2$a5÷a2 |
$=$= | $a^{5-2}$a5−2 |

$=$= | $a^3$a3 |

Let's look at another specific example. Say we wanted to find the value of $2^7\div2^3$27÷23. By evaluating each term in the quotient separately we would have

$2^7\div2^3$27÷23 | $=$= | $128\div8$128÷8 |

$=$= | $16$16 |

Alternatively, by first expanding the terms in the original expression we can arrive at a simplified version of the expression on our way to the final value.

$2^7\div2^3$27÷23 | $=$= | $\left(2\times2\times2\times2\times2\times2\times2\right)\div\left(2\times2\times2\right)$(2×2×2×2×2×2×2)÷(2×2×2) |

$=$= | $2^4$24 | |

$=$= | $16$16 |

Notice in the second line we have identified that $2^7\div2^3=2^4$27÷23=24.

We can avoid having to write each expression in expanded form by using the division law (which is also known as the quotient law).

The division law

$\frac{a^m}{a^n}=a^{m-n}$`a``m``a``n`=`a``m`−`n`, where $a$`a` is any number,

That is, when dividing terms with a common base:

- Keep the same base
- Find the difference in the power.

We can also write the division law in the form:

$a^m\div a^n=a^{m-n}$`a``m`÷`a``n`=`a``m`−`n`.

As with using the multiplication (or product) law, we can **only** apply the division (or quotient) law to terms with the **same bases** (just like we can only add and subtract like terms in algebra). We **can** simplify $\frac{9^8}{9^3}$9893 because the numerator and denominator have the **same base**: $9$9.

Careful!

We **cannot** simplify $\frac{8^5}{7^3}$8573 because the two terms **do not** have the same base (one has a base of $8$8 and the other has a base of $7$7).

But like with multiplication we **can** simplify $\frac{8^5}{2^5}$8525 by thinking of it as $\left(\frac{8}{2}\right)^5=4^5$(82)5=45.

Dividing terms with the same index

For any base numbers $a$`a` and $b$`b` that both have $n$`n` as a power,

$a^n\div b^n=\left(a\div b\right)^n$`a``n`÷`b``n`=(`a`÷`b`)`n`

That is, when dividing terms with the same index:

- Find the quotient of the bases
- Keep the same power

Write $37^{30}\div37^{18}$3730÷3718 in simplest index form.

Fill in the blank to make the equation true.

$14^9\div\left(\editable{}\right)^9=7^9$149÷()9=79

What happens if we want to divide one term by another and when we perform the subtraction and we are left with a power of $0$0? For example,

$4^5\div4^5$45÷45 | $=$= | $4^{5-5}$45−5 |

$=$= | $4^0$40 |

To think about what value we can assign to the term $4^0$40, let's write this division problem as the fraction $\frac{4^5}{4^5}$4545. Since the numerator and denominator are the same, the fraction simplifies to $1$1. Notice that this will also be the case with $\frac{4^{20}}{4^{20}}$420420 or any expression where we are dividing like bases whose powers are the same.

So the result we arrive at by using index laws is $4^0$40, and the result we arrive at by simplifying fractions is $1$1. This must mean that $4^0=1$40=1.

There is nothing special about the number $4$4, so we can extend this observation to any other base. This result is summarised by the zero power law.

The zero power law

For any base number $a$`a`,

$a^0=1$`a`0=1

This says that taking the zeroth power of any number will always result in $1$1.

Evaluate $8^0$80.

Perform simple operations with indices.