Indices

Matsec Syllabus: Pure Mathematics Paper 1 Topic 1: Positive and negative rational indices:

Properties of Indices i.e. zero, negative and fractional;
Applying the laws of indices;
Powers of products and quotients;
Simplifying expressions e.g. \frac{3(1+x)^2+4(1+x)^{-1}}{2(1+x)}.

Introduction

In this section, we are going to study numbers of the form:

a^m

where the number a is called the base and the number m is called the index (or power or exponent).

When the index is a positive integer, it shows by how many times the base is multiplied by itself. Hence:

a^m=\underbrace{a\times a \times \dots \times a}_{m\text{ times}}

Example 1: Express the following numbers in index form:

(a) 10000

\begin{aligned}&10000\\=&10^4\end{aligned}\\

(b) 64

\begin{aligned}&64\\=&4^3\end{aligned}\\

(c) 121

\begin{aligned}&121\\=&11^2\end{aligned}\\

(d) 49

\begin{aligned}&64\\=&7^2\end{aligned}\\

\begin{aligned}\sqrt{4}=\sqrt{2^2}=2\end{aligned}.

Three Rules of Indices

We have the following three rules of indices which are very important to perform algebraic procedures with indices. Let a, m and p be any three numbers.

Rule 1:

\boxed{a^m a^p=a^{m+p}}

Rule 2:

\boxed{\frac{a^m}{a^p}=a^{m-p}}

Rule 3:

\boxed{(a^m)^p=a^{mp}}

In Rule 1, if we have a product of two indices having the same base, the powers are added. In Rule 2, if we have a fraction of two indices having the same base, the power of the denominator is subtracted from the power of the numerator. In Rule 3, if a number in index form is again to the power of another number, the two powers are multiplied.

We are going to use the 3 rules in order to simplify expressions involving indices.

Example 2: Simplify the following:

(a) 4^2\times 4^3

\begin{aligned}&4^2\times 4^3\\=&4^5\text{ (by Rule 1)}\end{aligned}

(b) 3^{-3} 3^8

\begin{aligned}&3^{-3} 3^8\\=&3^5\text{ (by Rule 1)}\end{aligned}

(c) \frac{3^4}{3^2}

\begin{aligned}&\frac{3^4}{3^2}\\=&3^{4-2}\text{ (by Rule 2)}\\=&3^2\\=&9\end{aligned}

(d) \frac{2^3}{2^{-2}}

\begin{aligned}&\frac{2^3}{2^{-2}}\\=&2^{3-(-2)}\text{ (by Rule 2)}\\=&2^5\\=&32\end{aligned}

(e) (2^3)^2

\begin{aligned}&(2^3)^2\\=&2^6\text{ (by Rule 3)}\\=&64\end{aligned}

(f) (3^{-2})^{-2}

\begin{aligned}&(3^{-2})^{-2}\\=&3^4\text{ (By Rule 3)}\\=&81\end{aligned}

Power of Zero & Negative Power

From Rule 2, when m=p, it follows that:

\begin{aligned}\frac{a^m}{a^p}&=a^{m-p}\text{ (from Rule 2)}\\\frac{a^p}{a^p}&=a^{p-p}\text{ (letting }m=p)\\1&=a^0\end{aligned}

Let us call this Result 1.

Result 1:

\boxed{a^0=1}

Hence Result 1 tells us that any number to the power of zero, is equal to 1.

From Rule 2, when m=0, it follows that:

\begin{aligned}\frac{a^m}{a^p}&=a^{m-p}\text{ (from Rule 2)}\\\frac{a^0}{a^p}&=a^{0-p}\text{ (letting }m=0)\\\frac{1}{a^p}&=a^{-p}\text{ (since }a^0=1)\end{aligned}

Let us call this Result 2.

Result 2:

\boxed{a^{-p}=\frac{1}{a^p}}\text{}

Hence Result 2 tells us that whenever there is a negative sign in the power, this can be removed by performing the reciprocal of the number in index form.

We are going to use these 2 results in order to simplify expressions involving indices.

Example 3: Simplify the following:

(a) 5^0

\begin{aligned}&5^0\\=&1\text{ (by Result 1)}\end{aligned}

(b) (-2)^0

\begin{aligned}&(-2)^0\\=&1\text{ (by Result 1)}\end{aligned}

(c) 3^{-2}

\begin{aligned}&3^{-2}\\=&\frac{1}{3^2}\text{ (by Result 2)}\\=&\frac{1}{8}\end{aligned}

(d) (-3)^{-2}

\begin{aligned}&(-3)^{-2}\\=&\frac{1}{(-3)^2}\text{ (By Result 2)}\\=&\frac{1}{(-3)(-3)}\\=&\frac{1}{9}\end{aligned}

Fractions in Powers

Suppose that we have a fraction of the form \frac{1}{p} in the power, that is, there is 1 in the numerator and a positive whole number (integer) p in the denominator. This indicates that we have to compute the p^{th} root.

Rule 4:

\boxed{a^{\frac{1}{p}}=\sqrt[p]{a}}

Example 4: Simplify the following:

(a) 4^{\frac{1}{2}}

\begin{aligned}&4^{\frac{1}{2}}\\=&\sqrt[2]{4}\text{ (by Rule 4)}\\=&2\end{aligned}

(b) 27^{\frac{1}{3}}

\begin{aligned}&27^{\frac{1}{3}}\\=&\sqrt[3]{27}\text{ (by Rule 4)}\\=&3\end{aligned}

(c) (-8)^{\frac{1}{3}}

\begin{aligned}&(-8)^{\frac{1}{3}}\\=&\sqrt[3]{-8}\text{ (by Rule 4)}\\=&-2\end{aligned}

(d) 16^{0.25}

\begin{aligned}&16^{0.25}\\=&16^{\frac{1}{4}}\\=&\sqrt[4]{16}\text{ (by Rule 4)}\\=&2\end{aligned}

Now suppose that we have a fraction of the form \frac{m}{p} in the power, where m and p are positive whole numbers (integers). This indicates that the base is to make the base to the power of m and then take its p^{th} root. Or else, we switch the order… we take the p^{th} of the base and make it to the power of m.

Rule 5:

\boxed{a^{\frac{m}{p}}=\sqrt[p]{a^m}}\text{ or }\boxed{a^{\frac{m}{p}}={(\sqrt[p]{a})}^m}

Example 5: Simplify the following:

(a) 4^{\frac{3}{2}}

\begin{aligned}&4^{\frac{3}{2}}\\=&{\sqrt[2]{4}}^3\text{ (by Rule 5)}\\=&2^3\\=&8\end{aligned}

(b) 81^{0.75}

\begin{aligned}&81^{0.75}\\=&81^{\frac{3}{4}}\\=&{\sqrt[4]{81}}^3\text{ (by Rule 5)}\\=&3^3\\=&27\end{aligned}

(a) 4^{\frac{1}{2}} to continue

\begin{aligned}&4^{\frac{1}{2}}\\=&\sqrt[2]{4}\text{ (by Rule 4)}\\=&2\end{aligned}

(b) 27^{\frac{1}{3}} to continue

\begin{aligned}&27^{\frac{1}{3}}\\=&\sqrt[3]{27}\text{ (by Rule 4)}\\=&3\end{aligned}