MathsICSE Class 9 · Chapter 6

Indices (Exponents) ICSE Class 9: Laws of Indices with Solved Questions

Updated 15 October 2026 · Fact-checked

An index (exponent) tells how many times a base is multiplied by itself. ICSE Class 9 uses the laws aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ and (aᵐ)ⁿ = aᵐⁿ, with a⁰ = 1, a⁻ⁿ = 1/aⁿ and a^(1/n) as the nth root, to simplify expressions and solve equations such as 2ˣ⁺¹ = 32.

What Indices (Exponents) is about

In $a^{n}$ , $a$ is the base and $n$ is the index (also called the exponent or power). For a positive whole number $n$ , $a^{n}$ means $a \times a \times \dots \times a$ ( $n$ factors).

The CISCE 2028 Class IX syllabus asks you to handle positive, fractional, negative and zero indices, simplify expressions with various exponents, and use the three laws $a^{m} \times a^{n} = a^{m + n}$ , $a^{m} \div a^{n} = a^{m - n}$ and $(a^{m})^{n} = a^{mn}$ . The meanings of $a^{0}$ , $a^{- n}$ and $a^{\frac{p}{q}}$ all follow from these laws.

In the school exam you will evaluate numerical expressions such as $(\frac{16}{81})^{- \frac{3}{4}}$ , simplify algebraic ones, prove identities (for example that a product of indices equals 1), and solve exponential equations by making the bases equal. This chapter feeds straight into Logarithms, which come next.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Indices (Exponents) is in the 3. Algebra (iv) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

The three laws of indices

For any non-zero base $a$ and rational indices $m$ , $n$ :

1. Product law: $a^{m} \times a^{n} = a^{m + n}$ (same base: add indices).
2. Quotient law: $a^{m} \div a^{n} = a^{m - n}$ (same base: subtract indices).
3. Power law: $(a^{m})^{n} = a^{mn}$ (power of a power: multiply indices).

Two consequences used constantly: $(ab)^{n} = a^{n} b^{n}$ and $(\frac{a}{b})^{n} = \frac{a ^{n}}{b ^{n}}$ .

The laws work only for the same base (product and quotient) or the same index ( $(ab)^{n}$ ). $2^{3} \times 3^{2}$ cannot be combined into one power.

Worked example

Simplify $\frac{( x ^{3} ) ^{4} \times x ^{5}}{x ^{10}}$ .

Try it first, then show the solutionSolution

$(x^{3})^{4} = x^{12}$ .
$x^{12} \times x^{5} = x^{17}$ .
$x^{17} \div x^{10} = x^{7}$ .

Answer: $x^{7}$

Zero and negative indices

From the quotient law, $\frac{a ^{n}}{a ^{n}} = a^{n - n} = a^{0}$ ; but the fraction equals 1. So $a^{0} = 1$ for every $a \neq = 0$ .

Similarly $\frac{a ^{0}}{a ^{n}} = a^{- n}$ , so $a^{- n} = \frac{1}{a ^{n}}$ . A negative index means reciprocal, not a negative number: $2^{- 3} = \frac{1}{8}$ .

For fractions: $(\frac{a}{b})^{- n} = (\frac{b}{a})^{n}$ . Flip the fraction and make the index positive.

Worked example

Evaluate $5^{0} + 2^{- 2} + (\frac{2}{3})^{- 2}$ .

Try it first, then show the solutionSolution

$5^{0} = 1$ , $2^{- 2} = \frac{1}{4}$ , $(\frac{2}{3})^{- 2} = (\frac{3}{2})^{2} = \frac{9}{4}$ .
Sum $= 1 + \frac{1}{4} + \frac{9}{4} = 1 + \frac{10}{4} = \frac{7}{2}$ .

Answer: $\frac{7}{2}$

Fractional indices

$(a^{\frac{1}{n}})^{n} = a^{1} = a$ , so $a^{\frac{1}{n}}$ is the nth root of $a$ : $a^{\frac{1}{n}} = n a$ .

Then $a^{\frac{p}{q}} = (q a)^{p} = q a^{p}$ . Take the root first; the numbers stay small.

Example: $2 7^{\frac{2}{3}} = (327)^{2} = 3^{2} = 9$ .

With a negative fractional index, do three things: flip, root, power. $(\frac{16}{81})^{- \frac{3}{4}} = (\frac{81}{16})^{\frac{3}{4}} = (\frac{3}{2})^{3} = \frac{27}{8}$ .

Worked example

Evaluate $(64)^{- \frac{2}{3}} \times (\frac{1}{4})^{- \frac{3}{2}}$ .

Try it first, then show the solutionSolution

$6 4^{- \frac{2}{3}} = \frac{1}{( 3 64 ) ^{2}} = \frac{1}{16}$ .
$(\frac{1}{4})^{- \frac{3}{2}} = 4^{\frac{3}{2}} = (4)^{3} = 8$ .
Product $= \frac{8}{16} = \frac{1}{2}$ .

Answer: $\frac{1}{2}$

Simplifying with prime bases

When an expression mixes numbers like 4, 8, 9, 27, 125, write every number as a power of a prime: $4 = 2^{2}$ , $8 = 2^{3}$ , $9 = 3^{2}$ , $27 = 3^{3}$ , $125 = 5^{3}$ . Then apply the laws to each prime separately.

For sums like $2^{n} + 2^{n - 1}$ , take out the smallest power as a common factor: $2^{n - 1} (2 + 1) = 3 \cdot 2^{n - 1}$ . The laws do not let you add indices across a plus sign.

Worked example

Simplify $\frac{2 ^{n} + 2 ^{n - 1}}{2 ^{n + 1} - 2 ^{n}}$ .

Try it first, then show the solutionSolution

Numerator $= 2^{n - 1} (2 + 1) = 3 \cdot 2^{n - 1}$ .
Denominator $= 2^{n} (2 - 1) = 2^{n}$ .
Quotient $= \frac{3 \cdot 2 ^{n - 1}}{2 ^{n}} = 3 \cdot 2^{- 1} = \frac{3}{2}$ .

Answer: $\frac{3}{2}$

Exponential equations

If $a^{x} = a^{y}$ with $a > 0$ and $a \neq = 1$ , then $x = y$ . So:
1. Write both sides as powers of the same base.
2. Equate the indices.
3. Solve the resulting linear equation.

If different primes appear, as in $5^{x - 2} \times 3^{2 x - 3} = 135$ , write the right side in prime powers ( $135 = 5^{1} \times 3^{3}$ ) and equate the index of each prime.

If a power appears in two terms, such as $3^{x + 1} + 3^{x} = 108$ , take out the common power first: $3^{x} (3 + 1) = 108$ .

Worked example

Solve $4^{x + 1} = 8^{x - 1}$ .

Try it first, then show the solutionSolution

$(2^{2})^{x + 1} = (2^{3})^{x - 1}$ , so $2^{2 x + 2} = 2^{3 x - 3}$ .
$2 x + 2 = 3 x - 3$ .
$x = 5$ .

Answer: x = 5

Proving index identities

Questions ask you to show that a long product of powers equals 1, or that two expressions are equal. Strategy:

Combine each bracket into a single power of the base using the laws.
Add all the indices; if they total 0, the product is $x^{0} = 1$ .
Use identities such as $(a - b) (a + b) = a^{2} - b^{2}$ to simplify the indices.

For fractions like $\frac{1}{1 + x ^{a - b}}$ , multiply numerator and denominator by $x^{b}$ (or a suitable power) to get a common denominator.

Worked example

Show that $(\frac{x ^{a}}{x ^{b}})^{a + b} \times (\frac{x ^{b}}{x ^{c}})^{b + c} \times (\frac{x ^{c}}{x ^{a}})^{c + a} = 1$ .

Try it first, then show the solutionSolution

Product $= x^{(a - b) (a + b)} \times x^{(b - c) (b + c)} \times x^{(c - a) (c + a)}$ .
Total index $= (a^{2} - b^{2}) + (b^{2} - c^{2}) + (c^{2} - a^{2}) = 0$ .
So the product is $x^{0} = 1$ .

Answer: Equals 1, as required.

Maths in real life: Indices (Exponents)

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Repeated doubling

Suppose a bacterium splits into two every 20 minutes and you start with one bacterium.

How many bacteria are there after 4 hours, written as a power and as a number?

Try it first, then show the solutionSolution

4 hours $= 240$ minutes $= 12$ splitting periods.
Number $= 2^{12} = 4096$ .

Answer: 2¹² = 4,096 bacteria.

Memory hook: Doubling n times gives 2ⁿ.

Dividing powers of 10

Imagine light travels at $3 \times 1 0^{8}$ m/s and the Sun is taken to be $1.5 \times 1 0^{11}$ m away for this calculation.

How long does sunlight take to reach the Earth?

Try it first, then show the solutionSolution

Time $= \frac{1.5 \times 1 0 ^{11}}{3 \times 1 0 ^{8}} = 0.5 \times 1 0^{3}$ s.
$= 500$ s, a little over 8 minutes.

Answer: 500 seconds (about 8 min 20 s).

Memory hook: Divide the numbers, subtract the powers of 10.

Fractional index as a root

Suppose a cube-shaped tank holds $64$ m³ of water and a designer writes its edge as $6 4^{\frac{1}{3}}$ m.

Find the edge and the area of one face.

Try it first, then show the solutionSolution

Edge $= 6 4^{\frac{1}{3}} = 4$ m.
Face area $= 4^{2} = 16$ m², which is also $6 4^{\frac{2}{3}}$ .

Answer: Edge 4 m; face 16 m².

Memory hook: Power 1/3 undoes cubing.

Topics in this chapter

Indices (Exponents) formula sheet

Product law: $a^{m} \times a^{n} = a^{m + n}$; Same base only.
Quotient law: $a^{m} \div a^{n} = a^{m - n}$; a ≠ 0.
Power law: $(a^{m})^{n} = a^{mn}$; Multiply indices.
Zero index: $a^{0} = 1$; For a ≠ 0.
Negative index: $a^{- n} = \frac{1}{a ^{n}}$; (a/b)⁻ⁿ = (b/a)ⁿ.
Fractional index: $a^{\frac{p}{q}} = (q a)^{p}$; Root first, then power.
Same index: $(ab)^{n} = a^{n} b^{n}$; Also (a/b)ⁿ = aⁿ/bⁿ.

How the school exam asks Indices (Exponents)

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Evaluate a numerical expression	Find the value of expressions like 9^(5/2) − 3 × 5⁰ − (1/81)^(−1/2).	Handle each term separately: flip negative indices, take roots, then powers; combine at the end.
Simplify an algebraic expression	Simplify products and quotients of powers of x, a, b with fractional or negative indices.	Collect powers of each base; add or subtract indices; write the answer with positive indices.
Common-factor simplification	(2ⁿ + 2ⁿ⁻¹)/(2ⁿ⁺¹ − 2ⁿ) or (3ⁿ × 9ⁿ⁺¹)/(3ⁿ⁻¹ × 9ⁿ⁻¹).	Write in prime bases; take the smallest power out of sums.
Prove an identity	Show that a product of index expressions equals 1, or that two fractions with indices add to 1.	Add the indices; use algebraic identities in the indices.
Exponential equations	Solve 2^(x+1) = 32, 9^(x+2) = 720 + 9^x, 5^(x−2) × 3^(2x−3) = 135.	Same base on both sides; equate indices. For sums, take out the common power first.

Tricks to remember Indices (Exponents)

Short memory aids. Say each one aloud once, then use it on a question the same day.

Multiply → add, divide → subtract, power → multiply
The laws step down one level: multiplication of powers becomes addition of indices, and so on.
Use it when: Choosing the right law.
Negative means flip
A minus in the index flips the base; it never makes the answer negative.
Use it when: 2⁻³ = 1/8, not −8.
Root, then power
In a^(p/q), the bottom q is the root (it sits at the root of the fraction), the top p is the power.
Use it when: Fractional indices.
FRP
Negative fractional index: Flip, Root, Power.
Use it when: (16/81)^(−3/4).

Indices (Exponents): exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Value of a negative index · 1 mark · MCQExam level

The value of $(\frac{1}{8})^{- \frac{2}{3}}$ is:
(a) $- 4$
(b) $\frac{1}{4}$
(c) 4
(d) $- \frac{1}{4}$

Try it first, then show the solutionSolution

Flip: $8^{\frac{2}{3}}$ .
Root then power: $2^{2} = 4$ .

Answer: (c)

Example 2: Mixed evaluation · 3 marks · Short answerExam level

Evaluate $9^{\frac{5}{2}} - 3 \times 5^{0} - (\frac{1}{81})^{- \frac{1}{2}}$ .

Try it first, then show the solutionSolution

$9^{\frac{5}{2}} = 3^{5} = 243$ .
$3 \times 5^{0} = 3$ ; $(\frac{1}{81})^{- \frac{1}{2}} = 81 = 9$ .
$243 - 3 - 9 = 231$ .

Answer: 231

Example 3: Prime bases · 3 marks · Short answerExam level

Simplify $\frac{3 ^{n} \times 9 ^{n + 1}}{3 ^{n - 1} \times 9 ^{n - 1}}$ .

Try it first, then show the solutionSolution

Numerator $= 3^{n} \times 3^{2 n + 2} = 3^{3 n + 2}$ .
Denominator $= 3^{n - 1} \times 3^{2 n - 2} = 3^{3 n - 3}$ .
Quotient $= 3^{5} = 243$ .

Answer: 243

Example 4: Common power in a sum · 3 marks · Short answerHarder exam level

Solve $9^{x + 2} = 720 + 9^{x}$ .

Try it first, then show the solutionSolution

$9^{x} \times 81 - 9^{x} = 720$ .
$80 \times 9^{x} = 720$ , so $9^{x} = 9$ .
$x = 1$ .

Answer: x = 1

Example 5: Two primes · 3 marks · Short answerHarder exam level

Solve $5^{x - 2} \times 3^{2 x - 3} = 135$ .

Try it first, then show the solutionSolution

$135 = 5 \times 27 = 5^{1} \times 3^{3}$ .
Equate indices of 5: $x - 2 = 1$ , so $x = 3$ .
Check the index of 3: $2 (3) - 3 = 3$ ✓.

Answer: x = 3

Example 6: Fractions that add to 1 · 4 marks · Long answerHarder exam level

Prove that $\frac{1}{1 + x ^{a - b}} + \frac{1}{1 + x ^{b - a}} = 1$ .

Try it first, then show the solutionSolution

First term: multiply top and bottom by $x^{b}$ : $\frac{x ^{b}}{x ^{b} + x ^{a}}$ .
Second term: multiply top and bottom by $x^{a}$ : $\frac{x ^{a}}{x ^{a} + x ^{b}}$ .
Sum $= \frac{x ^{a} + x ^{b}}{x ^{a} + x ^{b}} = 1$ .

Answer: Equals 1, as required.

Example 7: Three equal powers · 4 marks · Long answerHOTS / competency

If $2^{x} = 3^{y} = 6^{- z}$ , show that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ .

Try it first, then show the solutionSolution

Let each equal $k$ . Then $2 = k^{\frac{1}{x}}$ , $3 = k^{\frac{1}{y}}$ , $6 = k^{- \frac{1}{z}}$ .
$6 = 2 \times 3 = k^{\frac{1}{x} + \frac{1}{y}}$ .
So $k^{\frac{1}{x} + \frac{1}{y}} = k^{- \frac{1}{z}}$ , giving $\frac{1}{x} + \frac{1}{y} = - \frac{1}{z}$ .
Hence $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ .

Answer: Shown.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Storage on a phone

Computer memory is measured in powers of 2. In this case study take 1 KB $= 2^{10}$ bytes, 1 MB $= 2^{10}$ KB and 1 GB $= 2^{10}$ MB. A phone has $2^{7}$ GB of storage. A song file takes $2^{22}$ bytes and a photo takes $2^{21}$ bytes.

(i) · 3 marks
Express the phone's storage in bytes as a power of 2.
Answer
Answer: $2^{37}$ bytes
$2^{7} \times 2^{10} \times 2^{10} \times 2^{10} = 2^{7 + 30} = 2^{37}$ .
(ii) · 3 marks
How many songs fit on the phone? Give the answer as a power of 2 and as a number.
Answer
Answer: $2^{15} = 32768$
$\frac{2 ^{37}}{2 ^{22}} = 2^{15} = 32768$ .
(iii) · 4 marks
If half the storage holds songs and the other half photos, how many photos fit? Show that this is twice the number of songs that fit in the same half.
Answer
Answer: $2^{15}$ photos
Half the storage $= 2^{36}$ bytes. Photos $= \frac{2 ^{36}}{2 ^{21}} = 2^{15} = 32768$ . Songs in that half $= \frac{2 ^{36}}{2 ^{22}} = 2^{14}$ . Ratio $= \frac{2 ^{15}}{2 ^{14}} = 2$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Evaluate $(81)^{\frac{3}{4}} - (\frac{1}{32})^{- \frac{2}{5}} + 8^{\frac{1}{3}} \times (\frac{1}{2})^{- 1} \times 2^{0}$ . (3 marks)

Step	Marks
$8 1^{\frac{3}{4}} = 3^{3} = 27$	1
$(\frac{1}{32})^{- \frac{2}{5}} = 3 2^{\frac{2}{5}} = 4$ ; $8^{\frac{1}{3}} \times 2 \times 1 = 4$	1
$27 - 4 + 4 = 27$	1

Simplify $(\frac{x ^{a + b}}{x ^{c}})^{a - b} \times (\frac{x ^{b + c}}{x ^{a}})^{b - c} \times (\frac{x ^{c + a}}{x ^{b}})^{c - a}$ . (4 marks)

Step	Marks
First factor: $x^{(a + b - c) (a - b)} = x^{a^{2} - b^{2} - a c + b c}$	1
Second: $x^{b^{2} - c^{2} - ab + a c}$ ; third: $x^{c^{2} - a^{2} - b c + ab}$	1
Total index: all terms cancel to 0	1
Answer $x^{0} = 1$	1

Solve for $x$ and $y$ : $4^{x + y} = 1024$ and $2 7^{x - y} = 3^{x + y - 2}$ . (5 marks)

Step	Marks
$4^{x + y} = 4^{5}$ , so $x + y = 5$	1
$3^{3 (x - y)} = 3^{x + y - 2}$	1
$3 x - 3 y = x + y - 2 \Rightarrow x - 2 y = - 1$	1
Subtract from $x + y = 5$ : $3 y = 6$ , $y = 2$	1
$x = 3$	1

Where students lose marks in Indices (Exponents)

The slips examiners see most often, and what to write instead.

Mistake: Treating a negative index as a negative number
Students write $2^{- 3} = - 8$ or $- \frac{1}{8}$ .
Wrong working
$2^{- 3} = - 8$
Right working
$2^{- 3} = \frac{1}{8}$
Do this: Negative index means reciprocal: flip, then apply the positive index.
Usually costs: The term and every step after it.
Mistake: Adding indices across a plus sign
$2^{n} + 2^{n - 1}$ is not $2^{2 n - 1}$ ; the product law needs multiplication.
Wrong working
$3^{x + 1} + 3^{x} = 3^{2 x + 1}$
Right working
$3^{x + 1} + 3^{x} = 3^{x} (3 + 1) = 4 \cdot 3^{x}$
Do this: Take out the smallest power as a common factor.
Usually costs: The whole question.
Mistake: Multiplying bases instead of keeping them
$2^{3} \times 2^{4}$ written as $4^{7}$ .
Wrong working
$2^{3} \times 2^{4} = 4^{7}$
Right working
$2^{3} \times 2^{4} = 2^{7}$
Do this: Same base: keep the base, add the indices.
Usually costs: 1 mark.
Mistake: Writing $a^{0} = 0$
Any non-zero number to the power 0 is 1, which is easy to forget inside a long expression.
Do this: Circle every zero index and replace it with 1 first.
Usually costs: 1 mark.
Mistake: Equating indices with different bases
$4^{x + 1} = 8^{x - 1}$ does not give $x + 1 = x - 1$ ; the bases differ.
Do this: Convert both sides to the same prime base before equating.
Usually costs: The whole equation.

Practice questions on Indices (Exponents)

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
$2^{- 2} \times 2^{5} \div 2^{0}$ equals:
(a) 8
(b) $\frac{1}{8}$
(c) 0
(d) 128
Hint
Add and subtract indices; 2⁰ = 1.
Answer
(a): $2^{3} = 8$ .
Q2 · 2 marks · Short answerExam level
Evaluate $(\frac{16}{81})^{- \frac{3}{4}}$ .
Hint
Flip, root, power.
Answer
$(\frac{81}{16})^{\frac{3}{4}} = (\frac{3}{2})^{3} = \frac{27}{8}$ .
Q3 · 3 marks · Short answerExam level
Solve $2^{x + 1} = 32$ and $3^{x + 1} + 3^{x} = 108$ .
Hint
32 = 2⁵; take out 3ˣ.
Answer
$x + 1 = 5$ , $x = 4$ . $4 \cdot 3^{x} = 108$ , $3^{x} = 27$ , $x = 3$ .
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): $(- 8)^{\frac{1}{3}} = - 2$ .
Reason (R): A negative index gives a negative value.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Is the index in (A) negative?
Answer
(a). $(- 2)^{3} = - 8$ , so the cube root of $- 8$ is $- 2$ ; (R) is false: a negative index gives a reciprocal.
Q5 · 3 marks · Short answerHarder exam level
Simplify $3 x^{- 6} y^{9} \times (x^{\frac{1}{2}} y^{- \frac{1}{4}})^{4}$ .
Hint
Cube root means power 1/3.
Answer
$x^{- 2} y^{3} \times x^{2} y^{- 1} = x^{0} y^{2} = y^{2}$ .
Q6 · 3 marks · Short answerHarder exam level
Simplify $\frac{5 ^{n + 3} - 6 \times 5 ^{n + 1}}{9 \times 5 ^{n} - 2 ^{2} \times 5 ^{n}}$ .
Hint
Take out 5ⁿ⁺¹ and 5ⁿ.
Answer
Numerator $= 5^{n + 1} (25 - 6) = 19 \cdot 5^{n + 1}$ ; denominator $= 5^{n} (9 - 4) = 5^{n + 1}$ . Answer 19.
Q7 · 3 marks · Short answerHarder exam level
Solve $(\frac{3}{5})^{x + 1} = \frac{125}{27}$ .
Hint
125/27 = (3/5)⁻³.
Answer
$(\frac{3}{5})^{\frac{x + 1}{2}} = (\frac{3}{5})^{- 3}$ , so $\frac{x + 1}{2} = - 3$ and $x = - 7$ .
Q8 · 4 marks · Long answerHarder exam level
If $a = 2^{\frac{1}{3}} + 2^{- \frac{1}{3}}$ , show that $2 a^{3} - 6 a = 5$ .
Hint
Cube a using (p + q)³ = p³ + q³ + 3pq(p + q), with pq = 1.
Answer
$a^{3} = 2 + \frac{1}{2} + 3 \cdot 1 \cdot a = \frac{5}{2} + 3 a$ . So $2 a^{3} = 5 + 6 a$ , i.e. $2 a^{3} - 6 a = 5$ .
Q9 · 4 marks · Long answerHarder exam level
Find the value of $\frac{9 ^{n} \times 3 ^{2} \times ( 3 ^{- \frac{n}{2}} ) ^{- 2} - 2 7 ^{n}}{3 ^{3 n} \times 8}$ .
Hint
Write everything as powers of 3.
Answer
$9^{n} \times 9 \times 3^{n} = 3^{3 n} \times 9$ ; numerator $= 3^{3 n} (9 - 1) = 8 \cdot 3^{3 n}$ . Value $= 1$ .
Q10 · 4 marks · Long answerHOTS / competency
Solve $2^{2 x} - 5 \times 2^{x} + 4 = 0$ .
Hint
Let y = 2ˣ.
Answer
$y^{2} - 5 y + 4 = 0 \Rightarrow (y - 1) (y - 4) = 0$ . $2^{x} = 1 \Rightarrow x = 0$ ; $2^{x} = 4 \Rightarrow x = 2$ .
Q11 · 4 marks · Long answerHOTS / competency
If $x = 3^{\frac{1}{3}} - 3^{- \frac{1}{3}}$ , find $3 x^{3} + 9 x$ .
Hint
Use (p − q)³ = p³ − q³ − 3pq(p − q), with pq = 1.
Answer
$x^{3} = 3 - \frac{1}{3} - 3 x$ , so $3 x^{3} = 8 - 9 x$ and $3 x^{3} + 9 x = 8$ .
Q12 · 3 marks · Short answerHarder exam level
Find the value of $(\frac{1}{8})^{- \frac{1}{3}} + (\frac{1}{27})^{- \frac{2}{3}} - (0.04)^{- \frac{1}{2}}$ .
Hint
0.04 = 1/25.
Answer
$2 + 9 - 5 = 6$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the product law of indices.Show answer
aᵐ × aⁿ = aᵐ⁺ⁿ.
2.What is a⁰ for a ≠ 0?Show answer
1.
3.What does a⁻ⁿ mean?Show answer
1/aⁿ.
4.What is a^(1/3)?Show answer
The cube root of a.
5.Evaluate 32^(3/5).Show answer
8.
6.Simplify (x²)³ ÷ x⁴.Show answer
x².
7.If 3ˣ = 81, what is x?Show answer
4.
8.Write 125 as a power of a prime.Show answer
5³.

Best order to study it

Learn the three laws with positive whole-number indices.
Derive and practise zero and negative indices.
Practise fractional indices: root first, then power.
Simplify mixed expressions using prime bases and common factors.
Finish with identities and exponential equations.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Do you add indices only when multiplying the same base?
Do you flip the base for a negative index?
Do you take the root before the power in a^(p/q)?
Do you convert 4, 8, 9, 27, 125 into prime powers?
Do you take out the smallest power from a sum of powers?
Can you show a product of powers equals 1 by adding indices?
Can you solve exponential equations by equating indices?

Indices (Exponents): frequently asked questions

Why is a⁰ equal to 1?

Because aⁿ ÷ aⁿ = 1, and by the quotient law it also equals aⁿ⁻ⁿ = a⁰. So a⁰ must be 1.

Is 0⁰ defined?

No, it is not defined in this course; the zero-index rule needs a ≠ 0.

Should the final answer have negative indices?

It is usual to write the answer with positive indices, for example 1/x³ rather than x⁻³, unless the question says otherwise.

Can I use a calculator for fractional indices?

Exam questions are set so that roots come out exactly (such as 81^(3/4) = 27), so you should not need one.

How are indices linked to logarithms?

A logarithm is an index: if aˣ = N, then log base a of N is x. The next chapter builds on these laws.

Based on: CISCE ICSE 2028 Mathematics (51) syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028