CBSE Class 8 Maths — Exponents and Powers

Chapter Overview & Weightage

Exponents and Powers is a foundational chapter in Class 8 Maths that prepares students for scientific notation, logarithms, and surds in higher classes. It bridges multiplication tables with the powerful concept of powers. In CBSE SA exams, this chapter carries 5–8 marks.

Year	Marks	Question Types
2023	6	2 MCQ + 1 SA (simplification)
2022	8	1 MCQ + 1 SA + 1 LA
2021	5	2 SA

Laws of exponents with integer bases (including negative exponents), simplification of expressions, and scientific notation problems appear in almost every CBSE Class 8 exam. Negative exponents and zero exponent are the most frequently tested concepts.

Key Concepts You Must Know

Exponent (power): In $a^n$ , $a$ is the base and $n$ is the exponent (or index or power). It means $a$ multiplied by itself $n$ times.

Negative exponent: $a^{-n} = \frac{1}{a^n}$

A negative exponent means “take the reciprocal and make the exponent positive.”

Zero exponent: $a^0 = 1$ for any $a \neq 0$ . Why? Because $\frac{a^n}{a^n} = a^{n-n} = a^0 = 1$ .

Laws of Exponents

$a^m \times a^n = a^{m+n}$ (same base: add exponents)
$\frac{a^m}{a^n} = a^{m-n}$ (same base: subtract exponents)
$(a^m)^n = a^{mn}$ (power of a power: multiply)
$(ab)^n = a^n b^n$ (power distributes over product)
$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ (power distributes over quotient)
$a^{-n} = \frac{1}{a^n}$ (negative exponent = reciprocal)
$a^0 = 1$ (any nonzero base, zero exponent = 1)

Scientific notation: A number written as $m \times 10^n$ where $1 \leq m < 10$ and $n$ is an integer.

$5,600,000 = 5.6 \times 10^6$ (large number → positive exponent)
$0.000034 = 3.4 \times 10^{-5}$ (small number → negative exponent)

Solved Previous Year Questions

PYQ 1 — 2023 CBSE

Q: Simplify: $\left(\frac{2}{3}\right)^{-2} \times \left(\frac{3}{4}\right)^2 \times \left(\frac{4}{2}\right)^0$

Step 1: Apply $a^{-n} = (1/a)^n = (a^{-1})^n$ :

\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4}

Step 2: $\left(\frac{3}{4}\right)^2 = \frac{9}{16}$

Step 3: Any number to the power 0 = 1: $\left(\frac{4}{2}\right)^0 = 1$

Step 4: $\frac{9}{4} \times \frac{9}{16} \times 1 = \frac{81}{64}$

Answer: $\frac{81}{64}$

PYQ 2 — 2022 CBSE

Q: Express in scientific notation: (a) 0.00000723 (b) 8,03,00,000

(a): Count decimal places moved to get 7.23: move 6 places right → $7.23 \times 10^{-6}$

(b): $8,03,00,000 = 8.03 \times 10^7$ (moved decimal 7 places left)

PYQ 3 — 2021 CBSE

Q: Find the value of $\left(\frac{1}{2}\right)^{-3} + \left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{4}\right)^{-1}$

\left(\frac{1}{2}\right)^{-3} = 2^3 = 8

\left(\frac{1}{3}\right)^{-2} = 3^2 = 9

\left(\frac{1}{4}\right)^{-1} = 4^1 = 4

= 8 + 9 - 4 = 13

Answer: 13

Difficulty Distribution

Difficulty	%	Types
Easy	40%	Basic law application, value of expressions
Medium	45%	Simplification using multiple laws, scientific notation
Hard	15%	Find unknown in exponential equations, comparing values

Expert Strategy

Learn the laws by doing, not memorising. The laws follow directly from the definition ( $a^m = a \times a \times ... \times a$ , $m$ times). Instead of memorising, derive them once and they’ll stick:

$a^m \times a^n = (a \times ... \times a) \times (a \times ... \times a) = a^{m+n}$

It’s just counting.

Negative exponent = flip the fraction. $(2/3)^{-4} = (3/2)^4$ . That’s all. Don’t write $1/(2/3)^4$ — it creates unnecessary complexity.

For expressions with multiple laws, simplify from the inside out. $(a^2)^3 \times a^4 = a^6 \times a^4 = a^{10}$ . Work step by step — don’t try to do two laws at once.

Finding unknowns in equations like $2^x = 32$ : Write 32 as a power of 2: $32 = 2^5$ , so $x = 5$ . Know powers of 2 up to $2^{10} = 1024$ and powers of 3 up to $3^5 = 243$ .

Common Traps

Trap 1: Writing $a^0 = 0$ instead of $a^0 = 1$ . Very common slip. Any nonzero base to the power zero is ALWAYS 1. Only $0^0$ is undefined.

Trap 2: Applying the product law when bases are different. $2^3 \times 3^3 \neq 6^6$ . The law $a^m \times a^n = a^{m+n}$ requires the SAME base. $2^3 \times 3^3 = (2 \times 3)^3 = 6^3$ using law 4, NOT $6^6$ .

Trap 3: Scientific notation with more than one digit before the decimal. $56 \times 10^4$ is NOT scientific notation. Standard form requires $1 \leq m < 10$ , so it should be $5.6 \times 10^5$ .

Trap 4: Handling negative bases with even/odd powers. $(-2)^4 = +16$ (positive — four negative signs = positive). $(-2)^3 = -8$ (negative — three negative signs = negative). Even power → always positive; odd power → sign of base.