Indices

UK Secondary (7-11)

Division law with integer bases

Lesson

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 $a^6\div a^2$`a`6÷`a`2, we could write it as:

We can see that there are six $a$`a`s being divided by two $a$`a`s to give a result of four $a$`a`s, and notice that $4$4 is the difference of the powers in the original expression.

So, in our example above,

$a^6\div a^2$a6÷a2 |
$=$= | $a^{6-2}$a6−2 |

$=$= | $a^4$a4 |

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.

Of course, 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).

Simplify the following by first writing it in expanded form: $\frac{3^7}{3^2}$3732.

**Think: **We want to first write the expression in expanded form, so that we can then cancel out common factors from the numerator and denominator.

**Do: **

$\frac{3^7}{3^2}$3732 | $=$= | $\frac{3\times3\times3\times3\times3\times3\times3}{3\times3}$3×3×3×3×3×3×33×3 |

$=$= | $\frac{3\times3\times3\times3\times3}{1}$3×3×3×3×31 | |

$=$= | $3^5$35 |

**Reflect:** We can see that the answer matches what we would expect if we used the division law. That is, $\frac{3^7}{3^2}=3^{7-2}$3732=37−2 giving us the final answer of $3^5$35.

We can also simplify expressions involving multiplication as well, such as $\frac{4\times3^6}{2\times3^4}$4×362×34. We follow the same steps as when we are multiplying expressions. That is, we can treat the problem in two parts. Let's take a look at an example.

Simplify $\left(8\times5^9\right)\div\left(2\times5^4\right)$(8×59)÷(2×54) using index laws.

**Think: **First, let's write the expression as a fraction. To make the simplifications easier, we can then split the fraction up into two fractions using the fact $\frac{a\times b}{c\times d}=\frac{a}{c}\times\frac{b}{d}$`a`×`b``c`×`d`=`a``c`×`b``d`.

**Do: **

$\left(8\times5^9\right)\div\left(2\times5^4\right)$(8×59)÷(2×54) | $=$= | $\frac{8\times5^9}{2\times5^4}$8×592×54 |

$=$= | $\frac{8}{2}\times\frac{5^9}{5^4}$82×5954 | |

$=$= | $4\times\frac{5^9}{5^4}$4×5954 | |

$=$= | $4\times5^5$4×55 | |

$=$= | $4\times5^5$4×55 |

**Reflect:** Combining the steps, we get $\left(8\times5^9\right)\div\left(2\times5^4\right)=4\times5^5$(8×59)÷(2×54)=4×55 and as this process becomes more familiar we can reduce the amount of intermediate steps we take to arrive at the solution.

Convert the following to fraction form and evaluate using index laws: $4^8\div4^5$48÷45.

Simplify the following, giving your answer in index form: $\frac{\left(-7\right)^{11}}{\left(-7\right)^5}$(−7)11(−7)5.

Evaluate the following using index laws: $5\times\frac{4^5}{4^3}$5×4543.