MathsICSE Class 8 · Exponents and Powers

Laws of Exponents with Integral (Positive, Zero and Negative) Powers: ICSE Class 8

Updated 17 October 2026 · Fact-checked

For a non-zero base $a$ and integers $m, n$: $a^m \times a^n = a^{m+n}$, $a^m \div a^n = a^{m-n}$, $(a^m)^n = a^{mn}$, $a^0 = 1$ and $a^{-n} = \frac{1}{a^n}$. For different bases with the same power, $a^n b^n = (ab)^n$ and $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$.

Notice the key action for each law: add, subtract, multiply, any a gives 1 for power 0, and a negative power flips the base.

Illustration drawn for QuestionBuddy.

Understand Laws of Exponents with Integral Powers

The CISCE curriculum limits Class VIII to integral powers, that is, powers that are whole numbers, zero or negative integers. In $a^{n}$ , $a$ is the base and $n$ the exponent.

**Laws (base $a \neq = 0$ , $m, n$ integers).**

Product law: $a^{m} \times a^{n} = a^{m + n}$ . Same base, so add the exponents.
Quotient law: $a^{m} \div a^{n} = a^{m - n}$ . Subtract the exponents.
Power of a power: $(a^{m})^{n} = a^{mn}$ . Multiply the exponents.
Zero exponent: $a^{0} = 1$ for every $a \neq = 0$ .
Negative exponent: $a^{- n} = \frac{1}{a ^{n}}$ . A negative exponent flips the base to the other side of the fraction bar; it does not make the number negative.
Same exponent: $a^{n} \times b^{n} = (ab)^{n}$ and $a^{n} \div b^{n} = (\frac{a}{b})^{n}$ .
Fractions: $(\frac{a}{b})^{- n} = (\frac{b}{a})^{n}$ .

Sign of a power. A negative base to an even power is positive; to an odd power it is negative: $(- 2)^{4} = 16$ , $(- 2)^{3} = - 8$ . Note $- 2^{4} = - 16$ , because the exponent acts only on $2$ .

Strategy for multi-step sums. Break every base into prime factors, apply the laws prime by prime, and only at the end multiply out. The laws apply only when the bases are the same (for adding exponents) or the exponents are the same (for combining bases). They never apply to a sum such as $2^{3} + 2^{2}$ .

Laws of Exponents with Integral Powers: formulas to remember

Product and quotient laws: $a^{m} \times a^{n} = a^{m + n}, a^{m} \div a^{n} = a^{m - n}$; Same non-zero base.
Power of a power: $(a^{m})^{n} = a^{mn}$; Multiply the exponents.
Zero and negative exponents: $a^{0} = 1, a^{- n} = \frac{1}{a ^{n}}$; $a \neq = 0$ .
Same exponent: $a^{n} b^{n} = (ab)^{n}, \frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}$; Combine bases when the exponents match.

How to solve Laws of Exponents with Integral Powers questions

Reduce every term to prime bases first, then use the laws one by one.

1Remove negative exponents by moving the base across the fraction bar.
2Replace any composite base by its prime factors, e.g. $12 = 2^{2} \times 3$ .
3Combine like bases with the product, quotient and power laws.
4Replace any $a^{0}$ by $1$ .
5Evaluate the remaining numerical powers and reduce the fraction.

Laws of Exponents with Integral Powers 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.

Repeated tripling

Suppose a video is shared so that every viewer sends it to 3 new people, and the first viewer is counted as round 0 with $3^{0} = 1$ viewer.

How many viewers are in round 5, and how many times larger is that than round 2?

Try it first, then show the solutionSolution

Round 5 has $3^{5} = 243$ viewers.
Ratio $= \frac{3 ^{5}}{3 ^{2}} = 3^{5 - 2} = 3^{3} = 27$ .

Answer: 243 viewers; 27 times larger.

Memory hook: Dividing powers of the same base subtracts the exponents.

Tricks to remember Laws of Exponents with Integral Powers

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

Add, subtract, multiply
Product law adds, quotient law subtracts, power-of-power multiplies exponents.
Use it when: Picking the right law.
Negative means flip
$a^{- n}$ is the reciprocal of $a^{n}$ ; it never changes the sign of the answer.
Use it when: Questions such as $(- 2)^{- 3}$ .

Laws of Exponents with Integral Powers: solved examples

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

Example 1: Product law · 1 mark · MCQBoard level

The value of $2^{- 3} \times 2^{5}$ is:
(a) $\frac{1}{4}$ (b) $4$ (c) $16$ (d) $64$

Try it first, then show the solutionSolution

Add exponents: $2^{- 3 + 5} = 2^{2} = 4$ .

Answer: (b)

Example 2: Product and quotient together · 2 marks · Short answerBoard level

Simplify $\frac{5 ^{7} \times 5 ^{- 3}}{5 ^{2}}$ .

Try it first, then show the solutionSolution

Numerator: $5^{7 + (- 3)} = 5^{4}$ .
Divide: $5^{4 - 2} = 5^{2} = 25$ .

Answer: 25

Example 3: Fraction with a negative power · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

$(\frac{2}{3})^{- 3} = (\frac{3}{2})^{3}$ .
$\frac{9}{4} = (\frac{3}{2})^{2}$ , so $(\frac{9}{4})^{2} = (\frac{3}{2})^{4}$ .
Product $= (\frac{3}{2})^{3 + 4} = \frac{3 ^{7}}{2 ^{7}} = \frac{2187}{128}$ .

Answer: $\frac{2187}{128}$

Example 4: Mixed laws with a zero power · 4 marks · Short answerHarder board level

Simplify $\frac{( - 3 ) ^{- 2} \times 6 ^{3}}{2 ^{- 1} \times 9 ^{0} \times 4 ^{2}}$ .

Try it first, then show the solutionSolution

$(- 3)^{- 2} = \frac{1}{9}$ ; $6^{3} = 216$ ; numerator $= \frac{216}{9} = 24$ .
$2^{- 1} = \frac{1}{2}$ ; $9^{0} = 1$ ; $4^{2} = 16$ ; denominator $= \frac{16}{2} = 8$ .
Value $= \frac{24}{8} = 3$ .

Answer: 3

Example 5: Independence of n · 4 marks · Long answerHOTS / competency

Show that $\frac{3 ^{n + 2} - 3 ^{n}}{3 ^{n + 1}}$ has the same value for every integer $n$ , and find it. Verify for $n = 1$ .

Try it first, then show the solutionSolution

Write $3^{n + 2} = 3^{n} \times 9$ and $3^{n + 1} = 3^{n} \times 3$ .
Numerator $= 3^{n} (9 - 1) = 8 \times 3^{n}$ .
Expression $= \frac{8 \times 3 ^{n}}{3 \times 3 ^{n}} = \frac{8}{3}$ , free of $n$ .
Check $n = 1$ : $\frac{27 - 3}{9} = \frac{24}{9} = \frac{8}{3}$ .

Answer: $\frac{8}{3}$ for every $n$ .

Where students lose marks in Laws of Exponents with Integral Powers

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

Mistake: Treating a negative exponent as a negative number
The exponent says "reciprocal", not "minus".
Wrong working
$2^{- 3} = - 8$ or $- 6$ .
Right working
$2^{- 3} = \frac{1}{2 ^{3}} = \frac{1}{8}$ .
Do this: Rewrite as a reciprocal first, then evaluate.
Usually costs: Answer mark and usually the whole line after it.
Mistake: Multiplying bases when adding exponents
The product law keeps the base and adds exponents.
Wrong working
$2^{3} \times 2^{2} = 4^{5}$ .
Right working
$2^{3} \times 2^{2} = 2^{5} = 32$ .
Do this: The base never changes under the product or quotient law.
Usually costs: The final mark.

Exam tips

Convert to prime bases at the start; it removes most errors in long fractions.
Keep the sign of the base inside brackets: $(- 3)^{2} = 9$ but $- 3^{2} = - 9$ .
Final answers: leave as a reduced fraction or whole number, not as a power, unless the question asks for a power.

Practice questions on Laws of Exponents with Integral 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 · MCQBoard level
$(- 2)^{- 3}$ equals:
(a) $\frac{1}{8}$ (b) $- \frac{1}{8}$ (c) $8$ (d) $- 8$
Hint
Flip the base, then deal with the sign of the odd power.
Answer
(b)
Q2 · 2 marks · Short answerBoard level
Evaluate $(3^{0} + 4^{- 1}) \times 2^{2}$ .
Hint
$3^{0} = 1$ and $4^{- 1} = \frac{1}{4}$ .
Answer
$(1 + \frac{1}{4}) \times 4 = \frac{5}{4} \times 4 = 5$ .
Q3 · 3 marks · Short answerHarder board level
Simplify $\frac{1 2 ^{4} \times 9 ^{3} \times 4}{6 ^{3} \times 8 ^{2} \times 27}$ .
Hint
Write every base in primes of 2 and 3.
Answer
Numerator: $2^{8} 3^{4} \cdot 3^{6} \cdot 2^{2} = 2^{10} 3^{10}$ . Denominator: $2^{3} 3^{3} \cdot 2^{6} \cdot 3^{3} = 2^{9} 3^{6}$ . Value $= 2^{1} \times 3^{4} = 162$ .
Q4 · 3 marks · Short answerHarder board level
Simplify $(\frac{x ^{- 2} y ^{3}}{x ^{3} y ^{- 1}})^{2}$ , writing the answer with positive exponents.
Hint
Simplify inside the bracket first.
Answer
Inside: $x^{- 2 - 3} y^{3 + 1} = x^{- 5} y^{4}$ . Square: $x^{- 10} y^{8} = \frac{y ^{8}}{x ^{10}}$ .
Q5 · 4 marks · Long answerHOTS / competency
Prove that $(a^{- 1} + b^{- 1})^{- 1} = \frac{ab}{a + b}$ and use it to evaluate $(2^{- 1} + 3^{- 1})^{- 1}$ .
Hint
Add the reciprocals as fractions before inverting.
Answer
$a^{- 1} + b^{- 1} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$ , so its inverse is $\frac{ab}{a + b}$ . For $a = 2, b = 3$ : $\frac{6}{5}$ (check: $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ , inverse $\frac{6}{5}$ ).

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 a^0 for a non-zero a?Show answer
1.
2.What is 5^-2 as a fraction?Show answer
1/25.
3.How do you simplify (a^m)^n?Show answer
a^(mn).
4.Can you combine 2^3 + 2^2 using a law?Show answer
No; the laws are for products and quotients only. 8 + 4 = 12.

Laws of Exponents with Integral Powers: frequently asked questions

Why is any non-zero number to the power 0 equal to 1?

Because a^m \div a^m = a^{m-m} = a^0, and any non-zero number divided by itself is 1.

Is 0^0 defined in Class 8?

No; the zero-power rule is only for non-zero bases.