This MCQ module is based on: Exponential Notation & Laws of Exponents
TOPIC 4 OF 22
Exponential Notation & Laws of Exponents
🎓 Class 8
Mathematics
CBSE
Theory
Ch 2 — Power Play (Exponents and Powers)
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Exponential Notation & Laws of Exponents
Targeting Class 8 level in Algebra, with Basic difficulty.
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Chapter 2: Power Play
Class 8 Mathematics — Ganita Prakash Part-I | Part 1 of 3: Exponential Notation & Operations
2.1 Experiencing the Power Play
Activity: The Paper Folding Challenge
Take a large sheet of paper. Fold it once, then again, and again. How many times can you fold it?
Most people find it impossible to fold a sheet more than 7–8 times by hand. But here is a stunning fact:
"If you could fold a sheet of paper 46 times, it would be so thick that it can reach the Moon!"
The initial thickness of a sheet of paper is about 0.001 cm. Each fold doubles the thickness. This is .
Paper Thickness After Folding
Thickness grows exponentially — each fold doubles the previous thickness
The table below shows the thickness at each fold. Notice how slow the growth is at first, then suddenly it becomes enormous:
| Fold | Thickness | Fold | Thickness | Fold | Thickness |
|---|---|---|---|---|---|
| 0 | 0.001 cm | 7 | 0.128 cm | 17 | ≈ 131 cm |
| 1 | 0.002 cm | 8 | 0.256 cm | 20 | ≈ 10.4 m |
| 2 | 0.004 cm | 9 | 0.512 cm | 27 | ≈ 1.3 km |
| 3 | 0.008 cm | 10 | 1.024 cm | 30 | ≈ 10.7 km |
| 4 | 0.016 cm | 11 | 2.048 cm | 36 | ≈ 687 km |
| 5 | 0.032 cm | 12 | 4.096 cm | 42 | ≈ 44,000 km |
| 6 | 0.064 cm | 13 | 8.192 cm | 46 | ≈ 7,00,000 km! |
Key Observation: Every 10 folds multiply the thickness by exactly 1024 (= 2¹⁰). After 30 folds the thickness is ~10.7 km (height planes fly). After just 46 folds, more than 7 lakh km — almost twice the distance to the Moon!
2.2 Exponential Notation and Operations
The thickness after each fold can be expressed neatly using powers:
From repeated multiplication to powers
After 1 fold: \(0.001 \times 2 = 0.002\) cm
After 2 folds: \(0.001 \times 2 \times 2 = 0.001 \times 2^2 = 0.004\) cm
After 3 folds: \(0.001 \times 2^3 = 0.008\) cm
After 7 folds: \(0.001 \times 2^7 = 0.128\) cm
After 46 folds: \(0.001 \times 2^{46}\) cm ≈ 7,00,000 km
What is Exponential Notation?
We already know \(n^2\) (n squared) and \(n^3\) (n cubed). Generalising:
- \(n \times n = n^2\) — "n raised to the power 2"
- \(n \times n \times n = n^3\) — "n raised to the power 3"
- \(n \times n \times n \times n = n^4\) — "n raised to the power 4"
- \(n \times n \times \ldots \times n \ (a\ \text{times}) = n^a\) — "n raised to the power a"
In 5⁴: base = 5, exponent = 4. Read as "5 raised to the power 4" or "4th power of 5"
Key vocabulary:
In \(5^4 = 625\):
- = 5 (the number being multiplied)
- = 4 (how many times to multiply)
- of 625 is \(5^4\)
More Examples
Worked Examples
\(4 \times 4 \times 4 = 4^3 = 64\)
\((-4) \times (-4) \times (-4) = (-4)^3 = -64\)
\(a \times a \times a \times b \times b = a^3b^2\) ("a cubed b squared")
\(a \times a \times b \times b \times b \times b = a^2b^4\) ("a squared b to the power 4")
Think About It
Note the difference between repeated addition and repeated multiplication:
\(4 + 4 + 4 = 3 \times 4 = 12\) ← linear growth (adds 4 each time)
\(4 \times 4 \times 4 = 4^3 = 64\) ← exponential growth (multiplies by 4 each time)
What is \(0^2\)? \(0^5\)? \(0^n\)? (Think: 0 multiplied by itself any number of times is always 0.)
Prime Factorisation in Exponential Form
Example: Express 32400 as a product of prime factors in exponential form
\(32400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 3 \times 3 \times 3 \times 3\)
\[32400 = 2^4 \times 5^2 \times 3^4\]
Law 1: Product of Powers with the Same Base
The Stones that Shine — A Counting Puzzle
Three daughters each got 3 baskets; each basket had 3 keys; each key opened 3 rooms; each room had 3 tables; each table had 3 necklaces; each necklace had 3 diamonds.
There are \(3^4 = 81\) rooms. Continuing: tables = \(3^5\), necklaces = \(3^6\), diamonds = \(3^7\).
How to compute \(3^7\) quickly? Notice:
\(3^7 = 3^4 \times 3^3 = 81 \times 27 = \mathbf{2187}\) diamonds
This leads us to the first law of exponents:
Law 1 — Product of Powers (same base):
\[n^a \times n^b = n^{a+b}\]where a and b are counting numbers.
Example: \(p^4 \times p^6 = p^{4+6} = p^{10}\)
Law 2: Power of a Power
Can \(4^6\) be computed in two different ways?
Both ways give 4096. Generalising: \(7^4 = (7^2)^2\) and \(2^{10} = (2^2)^5 = (2^5)^2\).
Law 2 — Power of a Power:
\[(n^a)^b = (n^b)^a = n^{a \times b} = n^{ab}\]where a and b are counting numbers.
Example: \((3^4)^2 = 3^8\) | \((2^3)^5 = 2^{15}\)
Law 3: Product of Powers with the Same Exponent
The Magical Pond
A lotus is placed in a doubling pond. After 4 days, the lotuses (now \(2^4\)) are moved to a tripling pond. After 4 more days:
\[\text{Lotuses} = 2^4 \times 3^4\]
Can this be expressed as a single power?
\[2^4 \times 3^4 = (2\times3) \times (2\times3) \times (2\times3) \times (2\times3) = (2\times3)^4 = 6^4 = 1296\]
Law 3 — Product with Same Exponent (different bases):
\[m^a \times n^a = (m \times n)^a\]where a is a counting number. Similarly: \(\dfrac{m^a}{n^a} = \left(\dfrac{m}{n}\right)^a\)
Example: \(2^5 \times 5^5 = (2 \times 5)^5 = 10^5 = 1{,}00{,}000\)
How Many Combinations?
Counting Combinations (Multiplication Principle)
Estu has 4 dresses and 3 caps. How many combinations?
For each of 3 caps, there are 4 dress choices → \(3 \times 4 = 12\) combinations.
Similarly, for a 5-digit PIN lock (digits 0–9 at each position):
\[10 \times 10 \times 10 \times 10 \times 10 = 10^5 = 1{,}00{,}000 \text{ passwords}\]Estu's 6-slot A–Z lock: \(26^6 = 3{,}08{,}91{,}57{,}576\) passwords — much safer!
Interactive: Exponent Explorer
Exponent Explorer
Figure it Out (Pages 22–23)
Question 1 — Express in exponential form
(i) \(6 \times 6 \times 6 \times 6\) (ii) \(y \times y\) (iii) \(b \times b \times b \times b\) (iv) \(5 \times 5 \times 7 \times 7 \times 7\) (v) \(2 \times 2 \times a \times a\) (vi) \(a \times a \times a \times c \times c \times c \times c \times d\)
(i) \(6^4\) (ii) \(y^2\) (iii) \(b^4\) (iv) \(5^2 \times 7^3\) (v) \(2^2 \times a^2\) (vi) \(a^3 c^4 d\)
Note for (vi): \(d = d^1\) — the exponent 1 is usually not written explicitly.
Question 2 — Prime factorisation in exponential form
(i) 648 (ii) 405 (iii) 540 (iv) 3600
(i) 648: \(648 = 8 \times 81 = 2^3 \times 3^4\)
(ii) 405: \(405 = 5 \times 81 = 5 \times 3^4 = 3^4 \times 5\)
(iii) 540: \(540 = 4 \times 135 = 4 \times 5 \times 27 = 2^2 \times 3^3 \times 5\)
(iv) 3600: \(3600 = 36 \times 100 = 2^2 \times 3^2 \times 2^2 \times 5^2 = 2^4 \times 3^2 \times 5^2\)
Question 3 — Numerical values
(i) \(2 \times 10^3\) (ii) \(7^2 \times 2^3\) (iii) \(3 \times 4^4\) (iv) \((-3)^2 \times (-5)^2\) (v) \(3^2 \times 10^4\) (vi) \((-2)^5 \times (-10)^6\)
(i) \(2 \times 1000 = 2000\)
(ii) \(49 \times 8 = 392\)
(iii) \(3 \times 256 = 768\)
(iv) \(9 \times 25 = 225\) [both negative × negative = positive]
(v) \(9 \times 10000 = 90{,}000\)
(vi) \((-32) \times 10^6 = -3{,}20{,}00{,}000\) [\((-2)^5 = -32\), \((-10)^6 = +10^6\)]
Competency-Based Questions
Q1. A new viral video gets shared 3 times by each viewer. Starting with 1 video shared on Day 0, how many total sharings happen by Day 5? Express your answer in exponential form and find its value. L1 Remember
Sharings on Day 5 = \(3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243\). This is exponential (multiplicative) growth — each sharing creates 3 more.
Q2. A school has 3 wings. Each wing has 3 floors. Each floor has 3 classrooms. Each classroom has 3 windows. Express the number of windows as a power of 3 and compute the value. L2 Understand
Windows = \(3 \times 3 \times 3 \times 3 = 3^4 = 81\).
Each "level" multiplies by 3, so after 4 levels of 3-branching, we get \(3^4\).
Q3. A smartphone PIN is 6 digits long, using only the digits 0–9 for the first 4 places, and only A–Z for the last 2 places. How many possible PINs are there? Express in exponential form. L3 Apply
First 4 places: \(10^4\) choices. Last 2 places: \(26^2\) choices.
Total = \(10^4 \times 26^2 = 10{,}000 \times 676 = 67{,}60{,}000\).
Q4. Priya says that \((2^3)^4\) and \((2^4)^3\) and \(2^{12}\) are all the same. Ravi says only two of them are equal. Who is correct? Justify with calculation. L4 Analyse
Priya is correct. By Law 2: \((2^3)^4 = 2^{3 \times 4} = 2^{12}\) and \((2^4)^3 = 2^{4 \times 3} = 2^{12}\).
\(2^{12} = 4096\). All three expressions equal 4096. ✓
Q5. Design a password system for a school ID card that uses a 4-symbol code. The first 2 symbols can be any digit (0–9), and the next 2 can be any uppercase letter (A–Z). Calculate the total number of possible codes. Is this system secure enough for 5,000 students? Explain. L5 Evaluate
Codes = \(10^2 \times 26^2 = 100 \times 676 = 67{,}600\).
For 5,000 students: \(67{,}600 \gg 5{,}000\), so there are more than 13 possible codes per student on average. This is reasonably secure for assignment. However, since students may guess codes, adding more symbols or mixing letters/digits would make it more robust.
Assertion–Reason Questions
Options: (A) Both Assertion and Reason are true; Reason is the correct explanation. (B) Both true; Reason is NOT the correct explanation. (C) Assertion is true, Reason is false. (D) Assertion is false, Reason is true.
ARQ 1.
Assertion: \(2^{10} = 1024\)
Reason: After every 10 folds of a paper, the thickness increases exactly 1024 times.
(A) — Both are true and the Reason correctly explains the Assertion. \(2^{10} = 1024\), and since folding doubles thickness, 10 folds → \(2^{10}\) times multiplication = 1024 times increase.
ARQ 2.
Assertion: \(m^a \times n^a = (mn)^a\)
Reason: This law applies because multiplication is commutative and associative.
(A) — Both true. \(m^a \times n^a = (m\times m\times\ldots)\times(n\times n\times\ldots) = (mn)\times(mn)\times\ldots = (mn)^a\). The regrouping relies on commutativity and associativity of multiplication.
ARQ 3.
Assertion: A 5-slot alphabetic lock (A–Z only) has fewer passwords than a 4-slot alphanumeric lock (digits + uppercase letters, i.e., 36 options per slot).
Reason: The number of options per slot determines the number of combinations, not the number of slots alone.
(A) — Alphabetic lock: \(26^5 = 1{,}18{,}81{,}376\). Alphanumeric lock: \(36^4 = 16{,}79{,}616\). So the 5-slot alphabetic lock actually has MORE combinations! Wait — Assertion is FALSE: \(26^5 > 36^4\).
Correct answer: (D) — Assertion is false (26⁵ = 11,881,376 > 36⁴ = 1,679,616). The Reason is true in general.
Frequently Asked Questions — Power Play (Exponents and Powers)
What is Exponential Notation & Laws of Exponents in NCERT Class 8 Mathematics?
Exponential Notation & Laws of Exponents is a key concept covered in NCERT Class 8 Mathematics, Chapter 2: Power Play (Exponents and Powers). This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Exponential Notation & Laws of Exponents step by step?
To solve problems on Exponential Notation & Laws of Exponents, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 2: Power Play (Exponents and Powers)?
The essential formulas of Chapter 2 (Power Play (Exponents and Powers)) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Exponential Notation & Laws of Exponents important for the Class 8 board exam?
Exponential Notation & Laws of Exponents is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Exponential Notation & Laws of Exponents?
Common mistakes in Exponential Notation & Laws of Exponents include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Exponential Notation & Laws of Exponents?
End-of-chapter NCERT exercises for Exponential Notation & Laws of Exponents cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.
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