[mathjax]
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 | am x an = am+n |
| Rule 2 | am ÷ an = am-n |
| Rule 3 | (am)n = amn |
| Rule 4 | a0=1 |
| Rule 5 | a-m = 1⁄am |
| Rule 6 | [mathjax] [latex]a^{\frac{m}{n}}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m[/latex] |
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 [latex]2^{3-k} = 8 \div 4^{1-k}[/latex], find the value of [latex]k[/latex].
[latex]2^{3-k} = 8 \div 4^{1-k}[/latex]
[latex]2^{3-k} = 2^3 \div (2^2)^{1-k}[/latex]
Apply Rule 3:
[latex]2^{3-k} = 2^3 \div 2^{2-2k}[/latex]
Apply Rule 2:
[latex]2^{3-k} = 2^{3-(2-2k)}[/latex]
Equate the indices
[latex]3-k = 3-2+2k[/latex]
[latex]2=3k[/latex]
[latex]k=\frac{2}{3}[/latex]
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!