The Laws of Indices are first introduced when a student goes to Secondary 3, though quite a handful of schools are teaching the topic in Secondary 2 and some schools will place this topic under one of the self-learning chapters (where schools use learning aid online and teachers will not do any formal teaching).

This topic serves as a foundation to Indices in Additional Mathematics (Amath) and will be a topic struggled by many students due to the weak foundation of Elementary Mathematics (Emath). Let’s take a look at the laws of Indices.

## laws of indices

Rule 1 | a^{m} x a^{n} = a^{m+n} |

Rule 2 | a^{m} ÷ a^{n} = a^{m-n} |

Rule 3 | (a^{m})^{n} = a^{mn} |

Rule 4 | a^{0}=1 |

Rule 5 | a^{-m} = ^{1}⁄_{am} |

Rule 6 | \(a^{\frac{m}{n}}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m\) |

The standard type of question for Emath will be indices presented in a product or quotient format, which we commonly called the **2 terms** question. Having recognise the **structure of the question** is the crux as it **comes naturally** to you and you can easily **make connections** to what you already know.

This is **Metacognitive Learning**. First, students have already considered what they know and acknowledge they have some **experience with similar** tasks. Next, things become evident in their head how they will apply their current knowledge to the new situation. Third, they know how to **fill the learning gaps **and **where to turn to** get the information they are lacking. We will want students to approach new learning where **students feel empowered**, **armed with a treasure chest of strategies** that help them to tackle new questions. This group of students will **start thinking before they start** the task and not blindly delve into the question.

__Commonwealth 2017 Prelim/EM/P1/Q6a__

A common mistake that we may see is that students use 8 divide 4 to get base 2. However, a check with the Rules of Indices reveal that this doesn’t belong to any of the rules. Instead, the trick here for solving 2 terms question is to make the **base same for each side**. The common base for 8 and 4 is 2. This is a standard procedure so that we can compare the index on both sides later. We shall do a detailed breakdown of the steps.

Given that \(2^{3-k} = 8 \div 4^{1-k}\), find the value of \(k\).

\(2^{3-k} = 8 \div 4^{1-k}\)

\(2^{3-k} = 2^3 \div (2^2)^{1-k}\)

**Apply Rule 3:
**\(2^{3-k} = 2^3 \div 2^{2-2k}\)

**Apply Rule 2:**

\(2^{3-k} = 2^{3-(2-2k)}\)

**Equate the indices**

\(3-k = 3-2+2k\)

\(2=3k\)

\(k=\frac{2}{3}\)

Having been very familiar with the task on hand is the key to getting tougher questions done at a later stage. I will share more about solving **3 terms Indices** question in AMath in my next post. Stay tuned!