O Level (Sec 3 & 4)Additional Mathematics
Indices
Master the laws of indices with this comprehensive study guide. Learn zero indices, negative indices, rational indices, and how to solve exponential equations. Includes 7 practice questions with step-by-step video solutions.
By Timothy Gan12 October 2021Singapore O-Level syllabus
Free revision pack
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- 9 concept sections
- 7 worked questions
- 16 video lessons
9
Concept sections
7
Practice questions
8
Concept videos
8
Solution videos
Start here
What Are Indices?
An index (or exponent) tells you the power to which a base is raised. In $a^n$, $a$ is the base and $n$ is the index. For a positive integer $n$, the notation means multiply $n$ copies of $a$:
$a^n=\underbrace{a\times a\times\cdots\times a}_{n\text{ factors}}$
For example, $3^4=3\times3\times3\times3=81$. It is read as '$3$ to the power of $4$'. Zero, negative and fractional indices extend this notation; they do not mean repeated multiplication in the same literal way. The laws below show how those powers behave.
Video lesson
Study Guide - Introduction to Index Notation
1
Concept
The 5 Laws of Indices
The laws of indices are a set of techniques to be applied when simplifying expressions containing powers. The purpose of the laws is to enable us to simplify problems of addition, subtraction, multiplication, and division involving powers. They are useful in many branches of mathematics, both for reducing lengthy calculations and for allowing us to work out a solution by inspection. There are $5$ important laws of indices.
In general, if $a$ is real number, and $m$ and $n$ are positive integers, then
Law 1 of Indices (same base):
${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
Example: ${{6}^{4}}\times {{6}^{5}}={{6}^{9}}$
Law 2 of Indices (same base):
$\frac{{a}^{m}}{{a}^{n}}={{a}^{m-n}}$, if $a \neq 0$.
$\frac{{a}^{m}}{{a}^{n}}={{a}^{m-n}}$, if $a \neq 0$.
Example: $\frac{{{10}^{6}}}{{{10}^{4}}}={{10}^{2}}$
Law 3 of Indices (same base):
${{({{a}^{m}})}^{n}}={{a}^{mn}}$
${{({{a}^{m}})}^{n}}={{a}^{mn}}$
Example: ${{\left( {{2}^{5}} \right)}^{2}}={{2}^{10}}$
If $a$ and $b$ are real numbers, and $n$ is a positive integer, then
Law 4 of Indices (same index):
${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$
${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$
Example: ${{\left( -3 \right)}^{2}}\times {{\left( -4 \right)}^{2}}={{\left( -3\times \left( -4 \right) \right)}^{2}}$
Law 5 of Indices (same index):
$\frac{{{a}^{m}}}{{{b}^{m}}}={{\left( \frac{a}{b} \right)}^{m}}$
$\frac{{{a}^{m}}}{{{b}^{m}}}={{\left( \frac{a}{b} \right)}^{m}}$
Example: $\frac{{{8}^{3}}}{{{5}^{3}}}={{\left( \frac{8}{5} \right)}^{3}}$
Video lesson
Study Guide - Laws of Indices
2
Concept
Zero Indices
Any number raised to the power zero will yield a value equal to one. In general, for $n=0$, we define
$a^0=1$
where $a$ is any real number and $a\neq0$.
Suppose we are required to simplify $\left( \frac{{{a}^{5}}}{{{a}^{5}}} \right)$,
Using Law 2 of Indices, $\left( \frac{{{a}^{5}}}{{{a}^{5}}} \right)={a}^{5-5}=a^0$,
But we also know that $\left( \frac{{{a}^{5}}}{{{a}^{5}}} \right)=1$
Therefore, $a^0=1$
Examples of Zero Indices:
$13^0=1$
$\left(-\frac{3}{7} \right)^0=1$
$5a^0=5(1)=5$
$13^0=1$
$\left(-\frac{3}{7} \right)^0=1$
$5a^0=5(1)=5$
Video lesson
Study Guide - Zero Indices
3
Concept
Negative Indices
The negative index notation is a convenient way to deal with situations where the power is a number less than $0$. What happens if we have a number with a negative power? Generally, we define it as
${a}^{-n}=\frac{1}{a^n}$
where $a$ is any real number and $a\neq 0$.
To simplify $\left( \frac{{{a}^{4}}}{{{a}^{6}}} \right)$,
We can use Law 2 of Indices, $\left( \frac{{{a}^{4}}}{{{a}^{6}}} \right)={a}^{4-6}={a}^{-2}$
But we also know that $\left( \frac{{{a}^{4}}}{{{a}^{6}}} \right)=\frac{1}{{a}^{2}}$
Therefore, ${a}^{-2}=\frac{1}{{a}^{2}}$
From this, we can derive: ${{\left( \frac{a}{b} \right)}^{-n}}={{\left( \frac{b}{a} \right)}^{n}}$
We can now extend all the $5$ laws of indices to include all integer indices (positive, zero, and negative).
Video lesson
Study Guide - Negative Indices
Video lesson
Extending the laws to zero and negative indices
4
Concept
Positive nth Root
An expression that involves the radical sign $\sqrt[n]{\square }$ is called the radical expression. The radical sign is a square root symbol. It is read as "root," and it means the principal square root. The number inside the radical sign is called the radicand.
If $a$ is a positive number such that $a={b}^{n}$ for some positive integer $b$, then $b$ is the positive ${n}^{th}$ root of $a$ and we write as:
$b=\sqrt[n]{a}$
Video lesson
Study Guide - Positive nth Root
5
Concept
Rational Indices
Since we have already extended all the 5 laws of indices to include all integer indices, is it possible to have an index that is not an integer? What happens if we encounter indices that are non-integer rational numbers?
If $a$ is a positive integer, we can define that
${a}^{\frac{1}{n}}=\sqrt[n]{a}$
if $a>0$.
Examples:
${7}^{\frac{1}{5}}=\sqrt[5]{7}$
${10}^{\frac{1}{11}}=\sqrt[11]{10}$
${7}^{\frac{1}{5}}=\sqrt[5]{7}$
${10}^{\frac{1}{11}}=\sqrt[11]{10}$
Generally, if $m$ and $n$ are positive integers,
${{a}^{\frac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ or ${{\left( \sqrt[n]{a} \right)}^{m}}$
if $a>0$.
Video lesson
Study Guide - Rational Indices
6
Concept
Common Indices Mistakes and How to Check Them
Use these quick checks before moving on from an indices question:
- Addition is not multiplication. $a^m+a^n$ cannot usually be simplified to $a^{m+n}$; that law only applies to a product.
- A negative index is not a negative value. $a^{-n}=\frac{1}{a^n}$, so move the factor across the fraction line and make the index positive.
- Apply an outside power to every factor. $(3x^2y)^2=9x^4y^2$, not $3x^4y^2$.
- Keep brackets around a negative base. $(-2)^4=16$, but $-2^4=-16$ because the power acts on $2$ before the negative sign.
- Reject impossible substitutions. If $u=a^x$ for $a>0$, then $u$ must be positive. A negative root from the substituted equation is invalid.
After simplifying, substitute a small non-zero value such as $a=2$ into the original and final expressions. Matching values do not prove every step, but a mismatch exposes an error quickly.
7
Concept
Equations Involving Indices
If we have an equation in which the unknown, $x$ is the index, we solve this by changing both sides to the same base. We can use prime factorization to break apart the numbers so that they have the same base.
For equations of the form $a^x=b$, express both sides with the same base and equate the powers.
Video lesson
Study Guide - Equations involving Indices
8
Concept
Solving Exponential Equations Using Substitution
Suppose you have to solve a complicated question with exponential equations, you can often make the question simpler by a process called substitution. The idea is to replace a hard-to-solve equation with an easier one.
It is called substitution because what gets substituted is a simpler expression that can then be solved using more basic math skills. You can substitute any expression for any variable in an equation as long as you don't change the value of whatever expression you're solving for.
9
Concept
Simultaneous Exponential Equations
In order to solve simultaneous exponential equations, we need to make sure that all base values in the equations are the same. Once both equations have the same base, we can equate the powers and solve the resulting linear simultaneous equations.
Guided practice
Practice Questions with Video Solutions
Attempt each question before opening the solution. Then compare your method with the worked steps and video explanation.
Question 12 video solutions
Simplifying Expressions with Laws of Indices
Question
Simplify each of the following:
(i) ${{a}^{7}}\times {{a}^{3}}$ ÷ ${{\left( {{a}^{3}} \right)}^{2}}$
(ii) ${{\left( {{2}^{b}} \right)}^{5}}$ ÷ $8{{b}^{2}}$
(iii) ${{\left( \frac{3{{x}^{2}}}{{{x}^{3}}} \right)}^{3}}$ ÷ $\frac{27{{x}^{7}}}{{{x}^{21}}}$
(iv) ${{\left( {{c}^{2}}d \right)}^{4}}\times {{\left( {{c}^{4}}{{d}^{3}} \right)}^{5}}$
(v) $5{{\left( ef \right)}^{3}}\times 10e{{f}^{2}}$
(vi) $16{{m}^{8}}{{n}^{7}}$ ÷ ${{\left( -2{{m}^{3}}{{n}^{2}} \right)}^{2}}$
(vii) ${{\left( \frac{{{p}^{2}}}{q} \right)}^{6}}\times {{\left( \frac{2{{q}^{2}}}{-3{{p}^{5}}} \right)}^{3}}$
(i) ${{a}^{7}}\times {{a}^{3}}$ ÷ ${{\left( {{a}^{3}} \right)}^{2}}$
(ii) ${{\left( {{2}^{b}} \right)}^{5}}$ ÷ $8{{b}^{2}}$
(iii) ${{\left( \frac{3{{x}^{2}}}{{{x}^{3}}} \right)}^{3}}$ ÷ $\frac{27{{x}^{7}}}{{{x}^{21}}}$
(iv) ${{\left( {{c}^{2}}d \right)}^{4}}\times {{\left( {{c}^{4}}{{d}^{3}} \right)}^{5}}$
(v) $5{{\left( ef \right)}^{3}}\times 10e{{f}^{2}}$
(vi) $16{{m}^{8}}{{n}^{7}}$ ÷ ${{\left( -2{{m}^{3}}{{n}^{2}} \right)}^{2}}$
(vii) ${{\left( \frac{{{p}^{2}}}{q} \right)}^{6}}\times {{\left( \frac{2{{q}^{2}}}{-3{{p}^{5}}} \right)}^{3}}$
Video lesson
Study Guide - Indices Question 1 Part 1
Video lesson
Part (2)
Step-by-step solution
- 1(i) ${{a}^{7}}\times {{a}^{3}}$ ÷ ${{\left( {{a}^{3}} \right)}^{2}} = {{a}^{7}}\times {{a}^{3}}$ ÷ ${{a}^{6}} = {a}^{7+3-6} = {a}^{4}$
- 2(ii) ${{\left( {{2}^{b}} \right)}^{5}}$ ÷ $8{{b}^{2}} = \frac{{{\left( 2b \right)}^{5}}}{8{{b}^{2}}} = \frac{{{2}^{5}}\cdot {{b}^{5}}}{8{{b}^{2}}} = \frac{{{2}^{5}}}{8}{{b}^{5-2}} = 4{{b}^{3}}$
- 3(iii) ${{\left( \frac{3{{x}^{2}}}{{{x}^{3}}} \right)}^{3}}$ ÷ $\frac{27{{x}^{7}}}{{{x}^{21}}} = \frac{{{3}^{3}}\cdot {{\left( {{x}^{2}} \right)}^{3}}}{{{\left( {{x}^{3}} \right)}^{3}}}\times \frac{{{x}^{21}}}{27{{x}^{7}}} = \frac{27{{x}^{6}}}{{{x}^{9}}}\times \frac{{{x}^{21}}}{27{{x}^{7}}} = \frac{{{x}^{6+21}}}{{{x}^{9+7}}} = {{x}^{27-16}} = {{x}^{11}}$
- 4(iv) ${{\left( {{c}^{2}}d \right)}^{4}}\times {{\left( {{c}^{4}}{{d}^{3}} \right)}^{5}} = {{\left( {{c}^{2}} \right)}^{4}}{{d}^{4}}\times {{\left( {{c}^{4}} \right)}^{5}}{{\left( {{d}^{3}} \right)}^{5}} = {{c}^{8}}{{d}^{4}}{{c}^{20}}{{d}^{15}} = {{c}^{28}}{{d}^{19}}$
- 5(v) $5{{\left( ef \right)}^{3}}\times 10e{{f}^{2}} = 5{{e}^{3}}{{f}^{3}}\times 10{{e}^{1}}{{f}^{2}} = 50{{e}^{3+1}}{{f}^{3+2}} = 50{{e}^{4}}{{f}^{5}}$
- 6(vi) $16{{m}^{8}}{{n}^{7}}$ ÷ ${{\left( -2{{m}^{3}}{{n}^{2}} \right)}^{2}} = \frac{16{{m}^{8}}{{n}^{7}}}{4{{m}^{6}}{{n}^{4}}} = 4{{m}^{2}}{{n}^{3}}$
- 7(vii) ${{\left( \frac{{{p}^{2}}}{q} \right)}^{6}}\times {{\left( \frac{2{{q}^{2}}}{-3{{p}^{5}}} \right)}^{3}} = \frac{{{p}^{12}}}{{{q}^{6}}}\times \frac{8{{q}^{6}}}{-27{{p}^{15}}} = -\frac{8}{27}{{p}^{-3}}$
Final answer
(i) $a^4$, (ii) $4b^3$, (iii) $x^{11}$, (iv) $c^{28}d^{19}$, (v) $50e^4f^5$, (vi) $4m^2n^3$, (vii) $-\frac{8}{27}p^{-3}$
Question 2Video solution
Simplifying with Zero and Negative Indices
Question
Simplify each of the following, giving your answers in positive index notation:
(i) $18{{a}^{-6}}$ ÷ $3{{\left( {{a}^{-2}} \right)}^{2}}$
(ii) $5{{b}^{0}}\times 3{{\left( {{b}^{-2}} \right)}^{2}}$
(iii) ${{\left( 3{{c}^{2}}{{d}^{-2}} \right)}^{2}}$
(iv) ${{\left( \frac{{{e}^{2}}{{f}^{-1}}}{2} \right)}^{-3}}$
(i) $18{{a}^{-6}}$ ÷ $3{{\left( {{a}^{-2}} \right)}^{2}}$
(ii) $5{{b}^{0}}\times 3{{\left( {{b}^{-2}} \right)}^{2}}$
(iii) ${{\left( 3{{c}^{2}}{{d}^{-2}} \right)}^{2}}$
(iv) ${{\left( \frac{{{e}^{2}}{{f}^{-1}}}{2} \right)}^{-3}}$
Video lesson
Study Guide - Indices Question 2
Step-by-step solution
- 1(i) $18{{a}^{-6}}$ ÷ $3{{\left( {{a}^{-2}} \right)}^{2}} = 18\left( \frac{1}{{{a}^{6}}} \right)\times \frac{1}{3{{a}^{-4}}} = 6\left( \frac{1}{{{a}^{6}}} \right)\times \frac{1}{{{a}^{-4}}} = \frac{6}{{{a}^{2}}}$
- 2(ii) $5{{b}^{0}}\times 3{{\left( {{b}^{-2}} \right)}^{2}} = 5\left( 1 \right)\times 3\left( {{b}^{-4}} \right) = 15{{b}^{-4}} = \frac{15}{{{b}^{4}}}$
- 3(iii) ${{\left( 3{{c}^{2}}{{d}^{-2}} \right)}^{2}} = {{3}^{2}}{{\left( {{c}^{2}} \right)}^{2}}{{\left( {{d}^{-2}} \right)}^{2}} = 9{{c}^{4}}{{d}^{-4}} = \frac{9{{c}^{4}}}{{{d}^{4}}}$
- 4(iv) ${{\left( \frac{{{e}^{2}}{{f}^{-1}}}{2} \right)}^{-3}} = {{\left( \frac{2}{{{e}^{2}}{{f}^{-1}}} \right)}^{3}} = \frac{8}{{{e}^{6}}{{f}^{-3}}} = \frac{8{{f}^{3}}}{{{e}^{6}}}$
Final answer
(i) $\frac{6}{a^2}$, (ii) $\frac{15}{b^4}$, (iii) $\frac{9c^4}{d^4}$, (iv) $\frac{8f^3}{e^6}$
Question 3Video solution
Evaluating nth Roots
Question
Evaluate each of the following without the use of a calculator:
(i) $\sqrt[4]{16}$
(ii) $\sqrt[3]{\frac{27}{125}}$
(i) $\sqrt[4]{16}$
(ii) $\sqrt[3]{\frac{27}{125}}$
Video lesson
Study Guide - Indices Question 3
Step-by-step solution
- 1(i) $\sqrt[4]{16} = \sqrt[4]{{{2}^{4}}} = 2$
- 2(ii) $\sqrt[3]{\frac{27}{125}} = \sqrt[3]{\frac{{{3}^{3}}}{{{5}^{3}}}} = \frac{\sqrt[3]{{{3}^{3}}}}{\sqrt[3]{{{5}^{3}}}} = \frac{3}{5}$
Final answer
(i) $2$, (ii) $\frac{3}{5}$
Question 4Video solution
Rational Indices in Radical Form
Question
Rewrite each of the following in radical form and hence evaluate the results without using a calculator:
(i) ${{81}^{\frac{1}{4}}}$
(ii) ${{8}^{-\frac{1}{3}}}$
(i) ${{81}^{\frac{1}{4}}}$
(ii) ${{8}^{-\frac{1}{3}}}$
Video lesson
Study Guide - Indices Question 4
Step-by-step solution
- 1(i) ${{81}^{\frac{1}{4}}} = \sqrt[4]{81} = \sqrt[4]{{{3}^{4}}} = 3$
- 2(ii) ${{8}^{-\frac{1}{3}}} = {{\left( {{8}^{-1}} \right)}^{\frac{1}{3}}} = {{\left( \frac{1}{8} \right)}^{\frac{1}{3}}} = \frac{1}{\sqrt[3]{8}} = \frac{1}{\sqrt[3]{{{2}^{3}}}} = \frac{1}{2}$
Final answer
(i) $3$, (ii) $\frac{1}{2}$
Question 5Video solution
Solving Simple Exponential Equations
Question
Solve each of the equation:
(i) ${2}^{x}=8$
(ii) ${5}^{y}=\frac{1}{25}$
(iii) ${9}^{x}=27$
(i) ${2}^{x}=8$
(ii) ${5}^{y}=\frac{1}{25}$
(iii) ${9}^{x}=27$
Video lesson
Study Guide - Indices Question 5
Step-by-step solution
- 1(i) ${{2}^{x}}=8$ means ${{2}^{x}}={{2}^{3}}$, therefore $x=3$
- 2(ii) ${{5}^{y}}=\frac{1}{25}$ means ${{5}^{y}}={{5}^{-2}}$, therefore $y=-2$
- 3(iii) ${{9}^{z}}=27$ means ${{\left( {{3}^{2}} \right)}^{z}}={{3}^{3}}$, so ${{3}^{2z}}={{3}^{3}}$, therefore $2z=3$ and $z=\frac{3}{2}=1\frac{1}{2}$
Final answer
(i) $x=3$, (ii) $y=-2$, (iii) $z=1\frac{1}{2}$
Question 6Video solution
Exponential Equations with Substitution
Question
Use an appropriate substitution, or otherwise, solve ${7}^{2x+1}+20({7}^{x})=3$.
Video lesson
Study Guide - Indices Question 6
Step-by-step solution
- 1Rewrite the equation: ${{7}^{2x+1}}+20\left( {{7}^{x}} \right)=3$
- 2Simplify: ${{7}^{2x}}\cdot 7+20\left( {{7}^{x}} \right)=3$
- 3Further: $7{{\left( {{7}^{x}} \right)}^{2}}+20\left( {{7}^{x}} \right)-3=0$
- 4Let $u={{7}^{x}}$, then: $7{{u}^{2}}+20u-3=0$
- 5Factor: $\left( 7u-1 \right)\left( u+3 \right)=0$
- 6So $u=\frac{1}{7}$ or $u=-3$ (rejected, as ${{7}^{x}}>0$)
- 7Replace $u$ with ${7}^{x}$: ${{7}^{x}}=\frac{1}{7}={{7}^{-1}}$
- 8Therefore $x=-1$
Final answer
$x=-1$
Question 7Video solution
Simultaneous Exponential Equations
Question
Solve the simultaneous equations:
${{4}^{x}}\left( {{2}^{y}} \right)=\frac{{{2}^{11}}}{{{16}^{y}}}$
${{5}^{x}}\left( {{5}^{x-6y}} \right)=1$
${{4}^{x}}\left( {{2}^{y}} \right)=\frac{{{2}^{11}}}{{{16}^{y}}}$
${{5}^{x}}\left( {{5}^{x-6y}} \right)=1$
Video lesson
Study Guide - Indices Question 7
Step-by-step solution
- 1From equation (1): ${{4}^{x}}\left( {{2}^{y}} \right)=\frac{{{2}^{11}}}{{{16}^{y}}}$
- 2Convert to base 2: ${{\left( {{2}^{2}} \right)}^{x}}\cdot \left( {{2}^{y}} \right)=\frac{{{2}^{11}}}{{{\left( {{2}^{4}} \right)}^{y}}}$
- 3Simplify: ${{2}^{2x}}\cdot {{2}^{y}}=\frac{{{2}^{11}}}{{{2}^{4y}}}$
- 4Combine powers: ${{2}^{2x+y}}={{2}^{11-4y}}$
- 5Equate indices: $2x+y=11-4y$, so $2x=11-5y$ ... (3)
- 6From equation (2): ${{5}^{x}}\left( {{5}^{x-6y}} \right)=1$
- 7Simplify: ${{5}^{x+\left( x-6y \right)}}={{5}^{0}}$
- 8Equate indices: $x+x-6y=0$, so $2x=6y$ and $x=3y$ ... (4)
- 9Substitute (4) into (3): $2\left( 3y \right)=11-5y$
- 10Solve: $6y=11-5y$, so $11y=11$ and $y=1$
- 11From (4): $x=3(1)=3$
Final answer
$x=3, y=1$
Revision summary
Key Formulas to Remember
Law 1: Multiplication (Same Base)
${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
When multiplying powers with the same base, add the indices
Law 2: Division (Same Base)
$\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
When dividing powers with the same base, subtract the indices
Law 3: Power of a Power
${{({{a}^{m}})}^{n}}={{a}^{mn}}$
When raising a power to another power, multiply the indices
Law 4: Multiplication (Same Index)
${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$
When powers have the same index, you can multiply the bases
Law 5: Division (Same Index)
$\frac{{{a}^{m}}}{{{b}^{m}}}={{\left( \frac{a}{b} \right)}^{m}}$
When powers have the same index, you can divide the bases
Zero Index
$a^0=1$
Any number (except 0) raised to the power of 0 equals 1
Negative Index
${a}^{-n}=\frac{1}{a^n}$
A negative power means the reciprocal of the positive power
Fractional Index
${a}^{\frac{1}{n}}=\sqrt[n]{a}$
A fractional index represents a root
General Fractional Index
${{a}^{\frac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ or ${{\left( \sqrt[n]{a} \right)}^{m}}$
The numerator is the power, the denominator is the root
Identities and Results to Memorise
- $a^0 = 1$ (where $a \neq 0$)
- $a^1 = a$
- $a^{-1} = \frac{1}{a}$
- ${{\left( \frac{a}{b} \right)}^{-n}}={{\left( \frac{b}{a} \right)}^{n}}$
- $\sqrt[n]{a} = a^{\frac{1}{n}}$
- $\sqrt[n]{a^m} = a^{\frac{m}{n}}$
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