MathsICSE Class 9 · Indices (Exponents)

Laws of Indices with Zero, Negative and Fractional Exponents (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

For a ≠ 0: aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ and (ab)ⁿ = aⁿbⁿ. Also a⁰ = 1, a⁻ⁿ = 1/aⁿ and a^(p/q) = (q-th root of a) raised to p. Write numbers with prime bases, then apply the laws to combine exponents.

Each step down divides by 2, so 2⁰ = 1 and negatives give fractions; for a^(p/q) take the root first, then the power.

Illustration drawn for QuestionBuddy.

Understand Laws of Indices: Zero, Negative and Fractional Exponents

The three laws. For a non-zero base $a$ and rational exponents $m, n$ :
$ $a^{m} \times a^{n} = a^{m + n}, \frac{a ^{m}}{a ^{n}} = a^{m - n}, (a^{m})^{n} = a^{mn} .$ $
Also $(ab)^{n} = a^{n} b^{n}$ and $(\frac{a}{b})^{n} = \frac{a ^{n}}{b ^{n}}$ .

Zero and negative exponents. From $\frac{a ^{m}}{a ^{m}} = a^{m - m} = a^{0}$ and $\frac{a ^{m}}{a ^{m}} = 1$ we get $a^{0} = 1$ ( $a \neq = 0$ ). From $a^{- n} \times a^{n} = a^{0} = 1$ we get
$ $a^{- n} = \frac{1}{a ^{n}}, (\frac{a}{b})^{- n} = (\frac{b}{a})^{n} .$ $
A negative exponent means "reciprocal", never a negative number: $2^{- 3} = \frac{1}{8}$ , not $- 8$ .

Fractional exponents. $a^{1/ q}$ is the $q$ -th root of $a$ and $a^{p / q} = (q a)^{p} = q a^{p}$ for $a > 0$ . Take the root first because it keeps numbers small: $2 7^{2/3} = (327)^{2} = 3^{2} = 9$ and $1 6^{- 3/4} = \frac{1}{( 4 16 ) ^{3}} = \frac{1}{8}$ .

Prime bases. Write every number as a product of prime powers ( $81 = 3^{4}$ , $\frac{125}{8} = (\frac{5}{2})^{3}$ ), then combine exponents of the same base.

Factor-and-cancel. For sums like $\frac{2 ^{n + 4} - 2 \cdot 2 ^{n}}{2 \cdot 2 ^{n + 3}}$ take out the smallest power: $\frac{2 ^{n} ( 16 - 2 )}{2 ^{n + 4}} = \frac{14}{16} = \frac{7}{8}$ . Never split $2^{n + 4} - 2^{n}$ as $2^{4}$ .

Proving identities. Reduce each side to a single power. For example
$(\frac{x ^{a}}{x ^{b}})^{a + b} (\frac{x ^{b}}{x ^{c}})^{b + c} (\frac{x ^{c}}{x ^{a}})^{c + a} = x^{(a^{2} - b^{2}) + (b^{2} - c^{2}) + (c^{2} - a^{2})} = x^{0} = 1.$

Level. Class IX is a school-exam class. Prime-base fraction questions are routine; the factor-and-cancel and proof items are at the hardest school-exam level.

Laws of Indices: Zero, Negative and Fractional Exponents: formulas to remember

Product: $a^{m} \cdot a^{n} = a^{m + n}$; Same base, add exponents.
Quotient: $\frac{a ^{m}}{a ^{n}} = a^{m - n}$; Subtract exponents.
Power of a power: $(a^{m})^{n} = a^{mn}$; Multiply exponents.
Zero and negative: $a^{0} = 1, a^{- n} = \frac{1}{a ^{n}}$; a ≠ 0.
Fractional: $a^{p / q} = (q a)^{p}$; Root first, then power.

How to solve Laws of Indices: Zero, Negative and Fractional Exponents questions

Convert to prime bases, apply one law at a time, and compute the arithmetic last.

1Write each number as a prime power (or a fraction of prime powers).
2Convert negative exponents by taking reciprocals.
3For fractional exponents, take the root first, then the power.
4Combine same-base terms using the product and quotient laws.
5If terms are added or subtracted, factor out the smallest power first.
6Simplify to a single number or a single power; check with a small value.

Laws of Indices: Zero, Negative and Fractional Exponents in real life

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.

Doubling culture

Suppose a lab culture starts with 500 cells and doubles every 3 hours, so after $t$ hours it has $500 \times 2^{t /3}$ cells (hypothetical numbers).

Find the number of cells at $t = 9$ , and at $t = - 3$ (three hours before the start).

Try it first, then show the solutionSolution

$t = 9$ : $500 \times 2^{3} = 4000$ .
$t = - 3$ : $500 \times 2^{- 1} = 250$ .

Answer: 4000 cells; 250 cells three hours earlier.

Memory hook: Negative exponents run the clock backwards.

Scaling a drawing

Imagine a designer scaling lengths by $(\frac{3}{2})^{- 2}$ in one step.

Find the scale factor.

Try it first, then show the solutionSolution

$(\frac{3}{2})^{- 2} = (\frac{2}{3})^{2} = \frac{4}{9}$ .

Answer: $\frac{4}{9}$

Memory hook: Flip the fraction, drop the minus.

Tricks to remember Laws of Indices: Zero, Negative and Fractional Exponents

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

Minus means flip
A negative exponent flips the base; it never makes the number negative.
Use it when: a⁻ⁿ questions.
Root then power
In a^(p/q), take the q-th root first.
Use it when: Keeping numbers small.
Prime bases
Rewrite as 2s, 3s, 5s and combine.
Use it when: All index simplifications.

Laws of Indices: Zero, Negative and Fractional Exponents: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Negative exponent · 1 mark · MCQBoard level

The value of $3^{- 2}$ is:
(a) $- 9$
(b) $- \frac{1}{9}$
(c) $\frac{1}{9}$
(d) $9$

Try it first, then show the solutionSolution

$3^{- 2} = \frac{1}{3 ^{2}} = \frac{1}{9}$ .

Answer: (c) $\frac{1}{9}$

Example 2: Fractional exponents · 3 marks · Short answerBoard level

Simplify $2 7^{2/3} \times 1 6^{- 3/4}$ .

Try it first, then show the solutionSolution

$2 7^{2/3} = (3^{3})^{2/3} = 3^{2} = 9$ .
$1 6^{- 3/4} = (2^{4})^{- 3/4} = 2^{- 3} = \frac{1}{8}$ .
Product $= \frac{9}{8}$ .

Answer: $\frac{9}{8}$

Example 3: Factor and cancel · 4 marks · Short answerHarder board level

Simplify $\frac{2 ^{n + 4} - 2 \cdot 2 ^{n}}{2 \cdot 2 ^{n + 3}}$ .

Try it first, then show the solutionSolution

Numerator: $2^{n + 4} - 2^{n + 1} = 2^{n} (2^{4} - 2) = 2^{n} \times 14$ .
Denominator: $2 \cdot 2^{n + 3} = 2^{n + 4} = 2^{n} \times 16$ .
Ratio $= \frac{14}{16} = \frac{7}{8}$ .

Answer: $\frac{7}{8}$

Example 4: Proof of an identity · 4 marks · Long answerHarder board level

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

Try it first, then show the solutionSolution

First bracket: $(x^{a - b})^{a + b} = x^{(a - b) (a + b)} = x^{a^{2} - b^{2}}$ .
Second: $x^{b^{2} - c^{2}}$ . Third: $x^{c^{2} - a^{2}}$ .
Product $= x^{(a^{2} - b^{2}) + (b^{2} - c^{2}) + (c^{2} - a^{2})} = x^{0} = 1$ . Hence proved.

Answer: Proved; the exponents add to 0.

Example 5: Mixed fractional and negative exponents · 4 marks · Short answerHOTS / competency

Evaluate $(\frac{125}{8})^{- 2/3} \times (\frac{81}{16})^{3/4}$ .

Try it first, then show the solutionSolution

$(\frac{125}{8})^{- 2/3} = (\frac{8}{125})^{2/3} = (\frac{2}{5})^{2} = \frac{4}{25}$ .
$(\frac{81}{16})^{3/4} = (\frac{3}{2})^{3} = \frac{27}{8}$ .
Product $= \frac{4}{25} \times \frac{27}{8} = \frac{27}{50}$ .

Answer: $\frac{27}{50}$

Where students lose marks in Laws of Indices: Zero, Negative and Fractional Exponents

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

Mistake: Negative answer from a negative exponent
It means reciprocal.
Wrong working
$2^{- 3} = - 8$ .
Right working
$2^{- 3} = \frac{1}{8}$ .
Do this: Reciprocal, then power.
Usually costs: Whole answer.
Mistake: Adding exponents of different bases
The product law needs the same base.
Wrong working
$2^{3} \times 3^{2} = 6^{5}$ .
Right working
$8 \times 9 = 72$ .
Do this: Check the bases.
Usually costs: Whole answer.
Mistake: Splitting a difference
$2^{n + 4} - 2^{n}$ is not $2^{4}$ .
Wrong working
$2^{n + 4} - 2^{n} = 2^{4}$ .
Right working
$2^{n} (16 - 1) = 15 \cdot 2^{n}$ .
Do this: Factor out the smallest power.
Usually costs: Whole answer.

Exam tips

Class IX is examined in school papers; exponent proofs usually collapse to x⁰ = 1.
Always rewrite in prime bases before combining.
For fractional exponents, take the root first.
Show each law you use by name (product, quotient, power of a power).

Practice questions on Laws of Indices: Zero, Negative and Fractional 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 · MCQBoard level
The value of $(\frac{1}{3})^{- 2}$ is:
(a) $\frac{1}{9}$
(b) $- \frac{1}{9}$
(c) $9$
(d) $- 9$
Hint
Flip and square.
Answer
(c) 9
Q2 · 3 marks · Short answerHarder board level
Evaluate $8^{2/3} + 2 5^{1/2} - 2 7^{- 1/3}$ .
Hint
Roots first.
Answer
$4 + 5 - \frac{1}{3} = \frac{26}{3}$ .
Q3 · 4 marks · Short answerHarder board level
Show that $(a^{- 1} + b^{- 1})^{- 1} = \frac{ab}{a + b}$ and find its value for $a = 2$ , $b = 3$ .
Hint
$a^{- 1} + b^{- 1} = \frac{a + b}{ab}$ .
Answer
$(\frac{a + b}{ab})^{- 1} = \frac{ab}{a + b}$ ; for $a = 2$ , $b = 3$ : $\frac{6}{5}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $2^{- 3} = - 8$ .
Reason (R): A negative exponent means the reciprocal of the positive power.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
$2^{- 3} = \frac{1}{8}$ .
Answer
(d)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A culture has $N = 500 \times 2^{t /3}$ cells after $t$ hours (hypothetical).
(a) Find $N$ at $t = 0$ .
(b) Find $N$ at $t = 9$ and $t = 6$ .
(c) Find $N$ at $t = - 3$ and explain the meaning.
Hint
Use 2^0, 2^3, 2^2, 2^(-1).
Answer
(a) 500 (b) 4000 and 2000 (c) 250: the number of cells 3 hours before the start.

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.What is a⁰ for a ≠ 0?Show answer
1.
2.What does a⁻ⁿ mean?Show answer
The reciprocal of aⁿ.
3.Write a^(p/q) using roots.Show answer
(q-th root of a) to the power p.
4.What is (aᵐ)ⁿ?Show answer
aᵐⁿ.

Laws of Indices: Zero, Negative and Fractional Exponents: frequently asked questions

Why is a⁰ = 1?

Because aᵐ ÷ aᵐ = a⁰ and also equals 1.

Is 0⁰ defined?

Class IX works with non-zero bases; 0⁰ is not used.

Do negative bases work with fractional exponents?

Class IX uses positive bases for fractional exponents.