MathsICSE Class 8 · Chapter 2

Exponents and Powers ICSE Class 8: Laws of Exponents with Integral Powers

Updated 15 October 2026 · Fact-checked

In aⁿ, a is the base and n is the exponent. ICSE Class 8 uses the laws aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ and (aᵐ)ⁿ = aᵐⁿ for any integers m and n, together with a⁰ = 1 and a⁻ⁿ = 1/aⁿ, to simplify expressions and to find an unknown exponent.

What Exponents and Powers is about

A power such as $2^{5}$ is a short way to write repeated multiplication: $2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$ . The number being multiplied (2) is the base and the small raised number (5) is the exponent or index.

The CISCE Class VIII curriculum asks for the laws of exponents with integral powers. That means the exponents can be positive whole numbers, zero or negative integers, so you must know what $5^{0}$ and $5^{- 2}$ mean as well as $5^{2}$ .

In a school exam you will:

evaluate expressions with negative and zero exponents,
simplify products and quotients by writing every number in prime factors,
find an unknown exponent from an equation such as $5^{m} \div 5^{- 3} = 5^{5}$ ,
work with powers of 10 for very large and very small quantities.

Fractional exponents (roots written as powers) are not part of Class VIII; they come in Class IX.

Weightage: CISCE publishes no question paper, marking scheme or chapter weightage for Class VIII Mathematics: your school sets and marks the exam. Exponents and Powers sits in the Number System theme of the CISCE Curriculum for Upper Primary Classes (VI-VIII).Source: CISCE Curriculum for Upper Primary Classes (VI-VIII), RDCD 2016

Key concepts

Base, exponent and the sign of a power

$a^{n}$ means $n$ factors of $a$ multiplied together, for a positive integer $n$ .

Watch the sign when the base is negative:

a negative base to an even power is positive: $(- 3)^{4} = 81$ ;
a negative base to an odd power is negative: $(- 2)^{5} = - 32$ .

Brackets matter. $(- 3)^{2} = 9$ , but $- 3^{2}$ means $- (3^{2}) = - 9$ , because the exponent belongs only to the 3.

Also $(- 1)^{n}$ is $1$ when $n$ is even and $- 1$ when $n$ is odd.

Worked example

Evaluate $(- 2)^{5}$ and $(- 3)^{4}$ .

Try it first, then show the solutionSolution

$(- 2)^{5} = (- 2) (- 2) (- 2) (- 2) (- 2)$ : five negative factors, an odd number, so the result is negative: $- 32$ .
$(- 3)^{4}$ : four negative factors, so the result is positive: $81$ .

Answer: $- 32$ and $81$

The three laws of exponents

For a non-zero base $a$ and integers $m$ , $n$ :

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

These laws hold for negative and zero exponents too, which is exactly what the Class VIII syllabus needs. They need the same base: $2^{3} \times 3^{2}$ cannot be joined into one power.

Worked example

Simplify $3^{5} \times 3^{- 2} \div 3^{4}$ .

Try it first, then show the solutionSolution

Product law: $3^{5} \times 3^{- 2} = 3^{5 + (- 2)} = 3^{3}$ .
Quotient law: $3^{3} \div 3^{4} = 3^{3 - 4} = 3^{- 1}$ .
$3^{- 1} = \frac{1}{3}$ .

Answer: $\frac{1}{3}$

Zero and negative exponents

Zero exponent: $\frac{a ^{n}}{a ^{n}} = 1$ , and by the quotient law it also equals $a^{n - n} = a^{0}$ . So $a^{0} = 1$ for every $a \neq = 0$ .

Negative exponent: $a^{- n} = \frac{1}{a ^{n}}$ . A negative exponent means take the reciprocal; it does not make the number negative. $2^{- 3} = \frac{1}{8}$ , not $- 8$ .

For a fraction, flip it and make the exponent positive: $(\frac{a}{b})^{- n} = (\frac{b}{a})^{n}$ . So $(\frac{2}{3})^{- 2} = (\frac{3}{2})^{2} = \frac{9}{4}$ .

Worked example

Evaluate $(\frac{2}{3})^{- 2} + (\frac{1}{2})^{- 3}$ .

Try it first, then show the solutionSolution

$(\frac{2}{3})^{- 2} = (\frac{3}{2})^{2} = \frac{9}{4}$ .
$(\frac{1}{2})^{- 3} = 2^{3} = 8$ .
Sum $= \frac{9}{4} + \frac{32}{4} = \frac{41}{4}$ .

Answer: $\frac{41}{4}$

Same exponent, different bases

When the exponents match, the bases can be combined:

$a^{n} \times b^{n} = (ab)^{n}$
$a^{n} \div b^{n} = (\frac{a}{b})^{n}$

This is handy for making powers of 10: $2^{4} \times 5^{4} = 1 0^{4} = 10000$ .

It also works in reverse: $6^{- 2} = 2^{- 2} \times 3^{- 2}$ and $1 0^{3} = 2^{3} \times 5^{3}$ , which is how you split composite bases into primes.

Worked example

Find $2^{5} \times 5^{5}$ without multiplying out each power.

Try it first, then show the solutionSolution

Same exponent: $2^{5} \times 5^{5} = (2 \times 5)^{5}$ .
$= 1 0^{5} = 100000$ .

Answer: $100000$

Simplifying with prime factors

For a mixed expression:
1. Write every number as a product of prime powers: $4 = 2^{2}$ , $8 = 2^{3}$ , $9 = 3^{2}$ , $10 = 2 \times 5$ , $25 = 5^{2}$ , $125 = 5^{3}$ .
2. Collect the powers of each prime separately, adding exponents on top and subtracting those underneath.
3. Rewrite any negative exponent as a reciprocal in the final answer.

This turns a frightening fraction into a short calculation.

Worked example

Simplify $\frac{16 \times 1 0 ^{2} \times 64}{2 ^{4} \times 4 ^{2}}$ .

Try it first, then show the solutionSolution

Numerator: $2^{4} \times 2^{2} \times 5^{2} \times 2^{6} = 2^{12} \times 5^{2}$ .
Denominator: $2^{4} \times 2^{4} = 2^{8}$ .
Quotient: $2^{12 - 8} \times 5^{2} = 2^{4} \times 25 = 16 \times 25$ .

Answer: $400$

Finding an unknown exponent

If $a^{x} = a^{y}$ (with $a \neq = 0, 1, - 1$ ), then $x = y$ . So to solve for an exponent:
1. Simplify each side to a single power of the same base.
2. Equate the exponents.
3. Solve the simple equation.

If the bases differ, change them first: $9 = 3^{2}$ , $\frac{1}{8} = 2^{- 3}$ , $27 = 3^{3}$ .

Worked example

Find $m$ if $5^{m} \div 5^{- 3} = 5^{5}$ .

Try it first, then show the solutionSolution

Quotient law: $5^{m - (- 3)} = 5^{m + 3}$ .
So $5^{m + 3} = 5^{5}$ , giving $m + 3 = 5$ .
$m = 2$ .

Answer: $m = 2$

Maths in real life: Exponents and Powers

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.

Positive powers as repeated multiplication

Suppose you send a message to 3 friends, and in each round every new receiver forwards it to 3 people who have not had it.

How many new people receive it in the 5th round?

Try it first, then show the solutionSolution

Round 1: $3$ people; round 2: $3^{2}$ ; each round multiplies by 3.
Round 5: $3^{5} = 243$ people.

Answer: $3^{5} = 243$ new people.

Memory hook: Multiplying by the same number again and again is a power.

Multiplying powers of 10

Imagine a cell that is $2 \times 1 0^{- 5}$ m wide is viewed under a microscope that magnifies $1 0^{3}$ times.

How wide does the cell appear?

Try it first, then show the solutionSolution

Image width $= 2 \times 1 0^{- 5} \times 1 0^{3} = 2 \times 1 0^{- 2}$ m.
$2 \times 1 0^{- 2}$ m $= 0.02$ m $= 2$ cm.

Answer: 2 cm

Memory hook: Add the exponents of 10: −5 + 3 = −2.

Negative powers as repeated halving

Suppose you take a 64 cm ribbon and cut it in half, keep one half, and repeat this 6 times.

How long is the piece you keep at the end?

Try it first, then show the solutionSolution

Each cut multiplies the length by $\frac{1}{2} = 2^{- 1}$ .
Length $= 64 \times 2^{- 6} = \frac{64}{64} = 1$ cm.

Answer: 1 cm

Memory hook: Halving n times multiplies by 2⁻ⁿ.

Topics in this chapter

Exponents and Powers formula sheet

Product law: $a^{m} \times a^{n} = a^{m + n}$; Same base: add exponents.
Quotient law: $a^{m} \div a^{n} = a^{m - n}$; a ≠ 0: subtract exponents.
Power law: $(a^{m})^{n} = a^{mn}$; Multiply exponents.
Zero exponent: $a^{0} = 1$; For any a ≠ 0.
Negative exponent: $a^{- n} = \frac{1}{a ^{n}}$; Also (a/b)⁻ⁿ = (b/a)ⁿ.
Same exponent: $a^{n} \times b^{n} = (ab)^{n}$; And aⁿ ÷ bⁿ = (a/b)ⁿ.

How the school exam asks Exponents and Powers

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

Question type	How it is asked	Method
Evaluate	Find the value of expressions such as (1/2)⁻² + (1/3)⁻² + (1/4)⁻².	Turn each negative exponent into a reciprocal, then add or multiply.
Simplify using laws	Simplify and write with positive exponents, e.g. (25 × t⁻⁴)/(5⁻³ × 10 × t⁻⁸).	Prime factors first; collect each base; finish with positive exponents.
Find the exponent	Find m so that (−3)ᵐ⁺¹ × (−3)⁵ = (−3)⁷.	Make one power of one base on each side and equate the exponents.
Find the multiplier or divisor	By what number should (−6)⁻¹ be multiplied to get 9⁻¹?	Required number = target ÷ given; evaluate both as fractions.
Compare powers	Arrange powers in order or decide which of two powers is larger.	Evaluate small ones; for large ones make the exponents equal and compare the bases.

Tricks to remember Exponents and Powers

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

Laws step down one level
Multiply powers → add exponents; divide powers → subtract exponents; power of a power → multiply exponents.
Use it when: Choosing the right law quickly.
Minus means flip
A minus sign in the exponent flips the base upside down. It never makes the answer negative.
Use it when: 2⁻³ = 1/8.
Anything to the zero is one
Any non-zero number to the power 0 is 1. Circle every zero exponent and write 1 straight away.
Use it when: Long expressions with a⁰.
Even hides the minus
An even power hides a negative base's minus sign; an odd power keeps it.
Use it when: Sign of (−a)ⁿ.

Exponents and Powers: exam-level solved questions

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

Example 1: Powers of −1 · 1 mark · MCQExam level

The value of $(- 1)^{51} + (- 1)^{50}$ is:
(a) $- 2$
(b) $0$
(c) $2$
(d) $1$

Try it first, then show the solutionSolution

51 is odd, so $(- 1)^{51} = - 1$ ; 50 is even, so $(- 1)^{50} = 1$ .
Sum $= - 1 + 1 = 0$ .

Answer: (b)

Example 2: Zero exponent inside an expression · 3 marks · Short answerExam level

Evaluate $(3^{- 1} + 4^{- 1} + 5^{- 1})^{0} + ((\frac{1}{2})^{- 1})^{2}$ .

Try it first, then show the solutionSolution

The first bracket is non-zero, so raised to the power 0 it equals 1.
$(\frac{1}{2})^{- 1} = 2$ , and $2^{2} = 4$ .
Total $= 1 + 4 = 5$ .

Answer: $5$

Example 3: Mixed bases with negative exponents · 3 marks · Short answerExam level

Simplify $(2^{- 1} \times 4^{- 1}) \div 2^{- 2}$ .

Try it first, then show the solutionSolution

$4^{- 1} = (2^{2})^{- 1} = 2^{- 2}$ .
$2^{- 1} \times 2^{- 2} = 2^{- 3}$ .
$2^{- 3} \div 2^{- 2} = 2^{- 3 + 2} = 2^{- 1} = \frac{1}{2}$ .

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

Example 4: Unknown exponent with a negative base · 3 marks · Short answerHarder exam level

Find $m$ so that $(- 3)^{m + 1} \times (- 3)^{5} = (- 3)^{7}$ .

Try it first, then show the solutionSolution

Product law: $(- 3)^{m + 1 + 5} = (- 3)^{m + 6}$ .
$(- 3)^{m + 6} = (- 3)^{7}$ , so $m + 6 = 7$ .
$m = 1$ .

Answer: $m = 1$

Example 5: Simplify with a variable · 4 marks · Long answerHarder exam level

Simplify $\frac{25 \times t ^{- 4}}{5 ^{- 3} \times 10 \times t ^{- 8}}$ ( $t \neq = 0$ ) and write the answer with positive exponents.

Try it first, then show the solutionSolution

Primes: $25 = 5^{2}$ and $10 = 2 \times 5$ .
Powers of 5: $5^{2} \div (5^{- 3} \times 5^{1}) = 5^{2 + 3 - 1} = 5^{4} = 625$ .
Powers of $t$ : $t^{- 4} \div t^{- 8} = t^{- 4 + 8} = t^{4}$ .
The 2 from the 10 stays underneath: answer $\frac{625 t ^{4}}{2}$ .

Answer: $\frac{625 t ^{4}}{2}$

Example 6: Equation in the exponent · 3 marks · Short answerHarder exam level

Find $x$ if $5^{2 x + 1} \div 25 = 125$ .

Try it first, then show the solutionSolution

$25 = 5^{2}$ and $125 = 5^{3}$ , so $5^{2 x + 1 - 2} = 5^{3}$ .
$2 x - 1 = 3$ .
$x = 2$ .

Answer: $x = 2$

Example 7: Folding paper · 5 marks · Long answerHOTS / competency

A sheet of paper is $1 0^{- 4}$ m thick. Each fold doubles the thickness. (i) How thick is the pile after 10 folds? (ii) What is the least number of folds that makes the pile thicker than 1 m?

Try it first, then show the solutionSolution

After $n$ folds the thickness is $2^{n} \times 1 0^{- 4}$ m.
(i) $2^{10} = 1024$ , so thickness $= 1024 \times 1 0^{- 4} = 0.1024$ m (about 10 cm).
(ii) Need $2^{n} \times 1 0^{- 4} > 1$ , that is $2^{n} > 1 0^{4} = 10000$ .
$2^{13} = 8192$ is too small; $2^{14} = 16384$ is enough.
So 14 folds are needed (thickness about 1.64 m).

Answer: (i) 0.1024 m; (ii) 14 folds

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.

Measuring very big and very small

Suppose a science club uses these units: 1 km $= 1 0^{3}$ m, 1 mm $= 1 0^{- 3}$ m and 1 micrometre (µm) $= 1 0^{- 6}$ m. Answer each part using the laws of exponents.

(i) · 3 marks
How many millimetres make one kilometre? Give the answer as a power of 10.
Answer
Answer: $1 0^{6}$
$\frac{1 0 ^{3}}{1 0 ^{- 3}} = 1 0^{3 - (- 3)} = 1 0^{6}$ mm.
(ii) · 3 marks
A virus is $1 0^{- 7}$ m long. How many such viruses placed end to end make 1 mm?
Answer
Answer: $1 0^{4} = 10000$
$\frac{1 0 ^{- 3}}{1 0 ^{- 7}} = 1 0^{- 3 + 7} = 1 0^{4}$ .
(iii) · 4 marks
Express 5 km in micrometres.
Answer
Answer: $5 \times 1 0^{9}$ µm
5 km $= 5 \times 1 0^{3}$ m. Number of µm $= \frac{5 \times 1 0 ^{3}}{1 0 ^{- 6}} = 5 \times 1 0^{3 + 6} = 5 \times 1 0^{9}$ .

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 $(4^{- 1} + 8^{- 1}) \div (\frac{2}{3})^{- 1}$ . (3 marks)

Step	Marks
$4^{- 1} + 8^{- 1} = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$	1
$(\frac{2}{3})^{- 1} = \frac{3}{2}$	1
$\frac{3}{8} \div \frac{3}{2} = \frac{3}{8} \times \frac{2}{3} = \frac{1}{4}$	1

Find $x$ if $2^{2 x - 1} = \frac{1}{8 ^{x - 3}}$ . (4 marks)

Step	Marks
$8 = 2^{3}$ , so the right side is $\frac{1}{2 ^{3 x - 9}}$	1
Negative exponent: $\frac{1}{2 ^{3 x - 9}} = 2^{- 3 x + 9}$	1
Equate exponents: $2 x - 1 = - 3 x + 9$	1
$5 x = 10$ , so $x = 2$	1

Simplify $\frac{3 ^{- 5} \times 1 0 ^{- 5} \times 125}{5 ^{- 7} \times 6 ^{- 5}}$ . (5 marks)

Step	Marks
Primes: $1 0^{- 5} = 2^{- 5} \times 5^{- 5}$ , $125 = 5^{3}$ , $6^{- 5} = 2^{- 5} \times 3^{- 5}$	1
Numerator $= 3^{- 5} \times 2^{- 5} \times 5^{- 5 + 3} = 3^{- 5} \times 2^{- 5} \times 5^{- 2}$	1
Denominator $= 5^{- 7} \times 2^{- 5} \times 3^{- 5}$	1
Powers of 2 and 3 cancel; $5^{- 2} \div 5^{- 7} = 5^{- 2 + 7} = 5^{5}$	1
$5^{5} = 3125$	1

Where students lose marks in Exponents and Powers

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

Mistake: Treating a negative exponent as a negative number
A negative exponent means reciprocal.
Wrong working
$5^{- 2} = - 25$
Right working
$5^{- 2} = \frac{1}{25}$
Do this: Flip the base, then use the positive exponent.
Usually costs: The term and everything that depends on it.
Mistake: Multiplying the bases when using the product law
The base stays the same; only the exponents are added.
Wrong working
$3^{2} \times 3^{4} = 9^{6}$
Right working
$3^{2} \times 3^{4} = 3^{6}$
Do this: Keep the base, add the exponents.
Usually costs: 1 mark.
Mistake: Writing $a^{0} = 0$
Any non-zero number to the power 0 is 1.
Wrong working
$(2^{- 1} + 3^{- 1})^{0} = 0$
Right working
$(2^{- 1} + 3^{- 1})^{0} = 1$
Do this: Replace every zero-exponent bracket by 1 before anything else.
Usually costs: 1 mark.
Mistake: Subtracting exponents the wrong way round
In $a^{m} \div a^{n}$ the exponent underneath is subtracted from the one on top.
Wrong working
$2^{3} \div 2^{5} = 2^{2}$
Right working
$2^{3} \div 2^{5} = 2^{3 - 5} = 2^{- 2} = \frac{1}{4}$
Do this: Always write top minus bottom: $m - n$ .
Usually costs: 1 to 2 marks.
Mistake: Losing the sign of a negative base
$(- 2)^{3}$ is negative, and $- 3^{2}$ is not the same as $(- 3)^{2}$ .
Do this: Count the negative factors: odd count gives a negative answer.
Usually costs: 1 mark.

Practice questions on Exponents and Powers

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
$5^{- 2}$ is equal to:
(a) $- 10$
(b) $- 25$
(c) $\frac{1}{25}$
(d) $\frac{1}{10}$
Hint
Negative exponent means reciprocal.
Answer
(c)
Q2 · 2 marks · Short answerExam level
Evaluate $(\frac{1}{2})^{- 2} + (\frac{1}{3})^{- 2} + (\frac{1}{4})^{- 2}$ .
Hint
Flip each fraction.
Answer
$4 + 9 + 16 = 29$ .
Q3 · 2 marks · Short answerExam level
Simplify $(- 4)^{5} \div (- 4)^{8}$ .
Hint
Subtract exponents, then watch the sign.
Answer
$(- 4)^{- 3} = \frac{1}{( - 4 ) ^{3}} = \frac{- 1}{64}$ .
Q4 · 3 marks · Short answerHarder exam level
Find $m$ if $9^{3} \times 3^{m} = 3^{10}$ .
Hint
Write 9 as a power of 3.
Answer
$3^{6} \times 3^{m} = 3^{10}$ , so $6 + m = 10$ and $m = 4$ .
Q5 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): $(- 2)^{- 4} = \frac{1}{16}$ .
Reason (R): An even power of a negative number is positive.
(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
$(- 2)^{- 4} = \frac{1}{( - 2 ) ^{4}}$ .
Answer
(c). $(- 2)^{4} = 16$ is positive because the power is even, so $(- 2)^{- 4} = \frac{1}{16}$ .
Q6 · 2 marks · Short answerHarder exam level
Evaluate $((\frac{1}{3})^{- 1} - (\frac{1}{4})^{- 1})^{- 1}$ .
Hint
Work inside the bracket first.
Answer
$(3 - 4)^{- 1} = (- 1)^{- 1} = - 1$ .
Q7 · 3 marks · Short answerHarder exam level
Simplify $\frac{( - 2 ) ^{- 3} \times 4 ^{2}}{8 ^{- 1}}$ .
Hint
$(- 2)^{- 3} = \frac{- 1}{8}$ .
Answer
$\frac{- 1}{8} \times 16 = - 2$ ; $- 2 \div \frac{1}{8} = - 16$ .
Q8 · 3 marks · Short answerHarder exam level
By what number should $(- 6)^{- 1}$ be multiplied so that the product is $9^{- 1}$ ?
Hint
Required number = $9^{- 1} \div (- 6)^{- 1}$ .
Answer
$\frac{1}{9} \div \frac{- 1}{6} = \frac{1}{9} \times (- 6) = \frac{- 2}{3}$ .
Q9 · 3 marks · Short answerHarder exam level
Find $x$ if $(\frac{5}{3})^{- 5} \times (\frac{5}{3})^{- 11} = (\frac{5}{3})^{8 x}$ .
Hint
Add the exponents on the left.
Answer
$- 16 = 8 x$ , so $x = - 2$ .
Q10 · 4 marks · Long answerHarder exam level
Simplify $\frac{2 ^{- 3} \times 3 ^{4} \times 5 ^{0}}{6 ^{- 2} \times 9}$ .
Hint
$6^{- 2} = 2^{- 2} \times 3^{- 2}$ and $9 = 3^{2}$ .
Answer
Denominator $= 2^{- 2} \times 3^{- 2} \times 3^{2} = 2^{- 2}$ . Value $= 2^{- 3 + 2} \times 3^{4} = \frac{81}{2}$ .
Q11 · 4 marks · Long answerHOTS / competency
Show that $(x^{a - b})^{c} \times (x^{b - c})^{a} \times (x^{c - a})^{b} = 1$ for $x \neq = 0$ and integers $a$ , $b$ , $c$ .
Hint
Add all the exponents.
Answer
Total exponent $= c (a - b) + a (b - c) + b (c - a) = a c - b c + ab - a c + b c - ab = 0$ . So the product is $x^{0} = 1$ .
Q12 · 3 marks · Short answerHOTS / competency
Without a calculator, decide which is greater: $2^{30}$ or $3^{20}$ . Explain.
Hint
Write both with exponent 10.
Answer
$2^{30} = (2^{3})^{10} = 8^{10}$ and $3^{20} = (3^{2})^{10} = 9^{10}$ . Same exponent, larger base, so $3^{20}$ is greater.

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.In 7⁴, which is the base and which the exponent?Show answer
Base 7, exponent 4.
2.State the quotient law.Show answer
aᵐ ÷ aⁿ = aᵐ⁻ⁿ (a ≠ 0).
3.What is (−7)⁰?Show answer
1.
4.Write 2⁻⁴ as a fraction.Show answer
1/16.
5.What is (3/4)⁻¹?Show answer
4/3.
6.Simplify (2³)⁻².Show answer
2⁻⁶ = 1/64.
7.Is (−5)⁷ positive or negative?Show answer
Negative (odd exponent).
8.Write 10⁴ × 10⁻⁶ as a power of 10.Show answer
10⁻².

Best order to study it

Revise positive powers and the sign rule for negative bases.
Learn the three laws with positive exponents, then extend to zero and negative exponents.
Practise evaluating expressions with reciprocals.
Simplify mixed expressions by writing numbers as prime powers.
Finish with equations in the exponent and powers of 10.

Quick revision checklist

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

Do you add exponents only when the bases are the same?
Do you flip the base for a negative exponent?
Do you replace a⁰ by 1 straight away?
Can you write 4, 8, 9, 25, 27 and 125 as prime powers?
Do you subtract exponents as top minus bottom?
Can you solve for an exponent by equating powers of one base?

Exponents and Powers: frequently asked questions

Why is a⁰ equal to 1 and not 0?

Because aⁿ ÷ aⁿ = 1, and the quotient law says it also equals aⁿ⁻ⁿ = a⁰. So a⁰ has to be 1 for the laws to stay true.

Is 0⁰ equal to 1?

No value is given to 0⁰ in this chapter. The zero-exponent rule needs a non-zero base.

Can a negative exponent give a negative answer?

Only if the base itself is negative and the exponent is odd, for example (−2)⁻³ = −1/8. The minus in the exponent alone only makes a reciprocal.

Should the final answer have negative exponents?

It is usual to give the answer with positive exponents, such as 1/x² instead of x⁻², unless the question asks otherwise.

Are fractional exponents like 4^(1/2) in Class 8?

No. The Class VIII syllabus uses integral (whole-number, zero and negative) exponents only. Fractional exponents come in Class IX.

Based on: CISCE Curriculum for Upper Primary Classes (VI-VIII), Mathematics (Class VIII); CISCE RDCD: Curriculum Pre-School to Class VIII