Indices

UK Secondary (7-11)

Simplify expressions using multiple index laws with integer bases

Lesson

We've learned a lot of index laws including the multiplication law, the division law, the power of a power law and the zero index law.

Here is a summary of the laws we know so far.

Summary of laws

The multiplication law: $a^m\times a^n=a^{m+n}$`a``m`×`a``n`=`a``m`+`n`

The division law: $a^m\div a^n=\frac{a^m}{a^n}=a^{m-n}$`a``m`÷`a``n`=`a``m``a``n`=`a``m`−`n`, $a\ne0$`a`≠0

The power of a power law: $\left(a^m\right)^n=a^{m\times n}$(`a``m`)`n`=`a``m`×`n`

The zero power law: $a^0=1$`a`0=1, $a\ne0$`a`≠0

Combining one or more of these laws, together with the order of operations, we can simplify more complicated expressions with negative indices (or powers). Let's look at some examples to see how.

Simplify the expression $8^2\times8^9\div8^3$82×89÷83.

**Think**: This expression contains three terms and two operations. We can perform one operation on two of the terms, then performing the remaining operation on the resulting expression.

**Do**: Let's begin with the multiplication, then do the division.

$8^2\times8^9\div8^3$82×89÷83 | $=$= | $8^{2+9}\div8^3$82+9÷83 |

$=$= | $8^{11}\div8^3$811÷83 | |

$=$= | $8^{11-3}$811−3 | |

$=$= | $8^8$88 |

**Reflect**: We combined the first two terms using the multiplication law. Once we had evaluated the new power we then used the division law to complete the simplification. Alternatively, we could have started by simplifying the last two terms using the division law, then completing the process with the multiplication law, as follows.

$8^2\times8^9\div8^3$82×89÷83 | $=$= | $8^2\times8^{9-3}$82×89−3 |

$=$= | $8^2\times8^6$82×86 | |

$=$= | $8^{2+6}$82+6 | |

$=$= | $8^8$88 |

So by the two approaches we get the same answer. This can be useful when one approach clearly involves less work than the other.

Simplify the expression $\frac{\left(20^3\right)^5}{\left(20^0\right)^7}$(203)5(200)7 using index laws.

**Think**: Right away we can see that simplifying this expression will involve the power of a power law and the zero power law. Once we complete the simplification we need to make sure our answer has a positive power.

**Do**:

$\frac{\left(20^3\right)^5}{\left(20^0\right)^7}$(203)5(200)7 | $=$= | $\frac{\left(20^3\right)^5}{1^7}$(203)517 | (Using the zero power law) |

$=$= | $\frac{\left(20^3\right)^5}{1}$(203)51 | ||

$=$= | $\left(20^3\right)^5$(203)5 | ||

$=$= | $20^{3\times5}$203×5 | (Using the power of a power law) | |

$=$= | $20^{15}$2015 |

**Reflect**: We could have used the power of a power law for both the numerator and the denominator from the beginning, which would give the step $\frac{20^{15}}{20^0}$2015200. And from there we could use the zero power law to simplify further. In this way we can be creative with the order that we use each law.

Simplify $10^6\times10^2\times10^9$106×102×109, giving your answer in index form.

Simplify $13^9\times13^2\div13^6$139×132÷136, giving your answer in index form.

Evaluate $\left(11^6\right)^5\div11^{30}$(116)5÷1130, by first simplifying using index laws.