Comprehensive worksheet designed to help students master the concepts of exponents and powers through various practice problems.

Let’s solve each problem step by step. We’ll go one question at a time, carefully calculating and simplifying.

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1. Find the value of each of the following

a. 13² = 13 × 13 = 169
b. 5³ = 5 × 5 × 5 = 25 × 5 = 125
c. 2⁴ = 2 × 2 × 2 × 2 = 4 × 4 = 16
d. 11² = 11 × 11 = 121
e. (–3)³ = (–3) × (–3) × (–3) = 9 × (–3) = –27
f. (–1)⁶ = (–1) × (–1) × (–1) × (–1) × (–1) × (–1) → even number of negatives → positive → 1

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2. Simplify

a. 3 × 10² = 3 × 100 = 300
b. 2⁵ × 5³ = (32) × (125) = let’s compute: 32 × 100 = 3200; 32 × 25 = 800 → total = 3200 + 800 = 4000
*(Alternatively: 2⁵ × 5³ = 2² × (2³ × 5³) = 4 × (10³) = 4 × 1000 = 4000)*
c. 0 × 10⁴ = 0 (anything times zero is zero)
d. (3/4)³ = (3³)/(4³) = 27 / 64 → 27/64
e. ((–2)/3)⁴ = (–2)⁴ / 3⁴ = 16 / 81 → since exponent is even, negative becomes positive → 16/81

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3. Express each in exponential form

a. (–5/7) × (–5/7) × (–5/7) × (–5/7) → four times → (–5/7)⁴
b. –5 × –5 × –5 → three times → (–5)³
c. x × x × x × x × x × a × a × b × b × b → count: x⁵, a², b³ → x⁵ a² b³
d. (–2) × (–2) × (–2) × (–2) × a × a × a → (–2)⁴ × a³ → (–2)⁴ a³

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4. Express as product of powers of prime factors

We break each number into primes:

a. 36 = 6×6 = (2×3)×(2×3) = 2² × 3² → 2² × 3²
b. 675 → divide by 5: 675 ÷ 5 = 135; ÷5 again = 27; 27 = 3³ → so 5² × 3³ → 3³ × 5²
c. 392 → even → ÷2=196; ÷2=98; ÷2=49; 49=7² → so 2³ × 7² → 2³ × 7²
d. 864 → ÷2=432; ÷2=216; ÷2=108; ÷2=54; ÷2=27; 27=3³ → so 2⁵ × 3³ → 2⁵ × 3³
e. 450 → ÷2=225; 225=15×15=(3×5)²=3²×5² → so 2 × 3² × 5² → 2 × 3² × 5²
f. 1800 → ÷100=18 → 100=2²×5²; 18=2×3² → total: 2³ × 3² × 5² → 2³ × 3² × 5²

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5. Using laws of exponents, simplify

Recall:
- a^m × a^n = a^(m+n)
- a^m ÷ a^n = a^(m-n)
- (a^m)^n = a^(m×n)
- a^0 = 1

(i) 3⁶ × 3² = 3^(6+2) = 3⁸
(ii) (7²)³ ÷ 7³ = 7^(2×3) ÷ 7³ = 7⁶ ÷ 7³ = 7^(6-3) = 7³
(iii) 2²⁰ ÷ 2⁵ = 2^(20-5) = 2¹⁵
(iv) 2⁴ × 5⁴ = (2×5)⁴ = 10⁴
(v) (2⁰ + 3⁰)(4⁰ + 6⁰) = (1 + 1)(1 + 1) = 2 × 2 = 4
(vi) 7² / 5² = (7/5)² → but if they want simplified fraction: 49/25 → but exponential form? Probably leave as (7/5)² or just 49/25 — but since it says “simplify”, and no base same, we can write as (7²)/(5²) or decimal? But likely keep as fraction: 49/25

Wait — looking back, part (vi) is written as:

> (vi) 7² / 5²

So simplifying means compute numerator and denominator? Or leave as power? Since others are using exponent rules, probably leave as (7/5)² or just state 49/25. But let’s check context — other parts use exponent laws. So perhaps they want to combine? But bases different. So best answer: 49/25

But actually, in many curricula, they accept (7/5)² as simplified exponential form. Let’s go with (7/5)² for consistency with exponent theme.

Actually, rechecking: the instruction says “simplify” — so numerical value might be expected. But 49/25 is fine. However, let’s see what’s standard.

Looking at (iv): 2⁴ × 5⁴ = 10⁴ — that’s combining. For (vi), since bases differ, cannot combine exponents. So simplest is to compute: 49/25.

I think both are acceptable, but since it's "simplify", and 49/25 is a single fraction, I'll go with 49/25

But wait — original has “7² / 5²” — so maybe they want it as (7/5)²? Let me check common practice.

In NCERT (Indian curriculum), often they leave as (7/5)² if asking for exponential form, but here it says “simplify”. Hmm.

To be safe, I’ll compute: 7² = 49, 5² = 25 → 49/25

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6. Simplify and express in exponential form

(i) 2¹⁵ / (2⁷ × 2³) = 2¹⁵ / 2^(7+3) = 2¹⁵ / 2¹⁰ = 2^(15-10) = 2⁵

(ii) (3⁵ × 3²)³ = (3^(5+2))³ = (3⁷)³ = 3^(7×3) = 3²¹

(iii) [(2³)⁴ × 2⁸] ÷ 2¹² = [2^(3×4) × 2⁸] ÷ 2¹² = [2¹² × 2⁸] ÷ 2¹² = 2^(12+8) ÷ 2¹² = 2²⁰ ÷ 2¹² = 2^(20-12) = 2⁸

(iv) (5⁴ × x¹⁰ y⁵) / (5⁴ × x y⁴) = cancel 5⁴ top and bottom → x^(10-7) y^(5-4) = x³ y

(v) (2/3)⁵ × (3/5)⁵ = [ (2/3) × (3/5) ]⁵ = (2/5)⁵ → because 3 cancels → (2/5)⁵

(vi) [9⁸ × (x²)⁵] / [ (27)⁴ × (x³)² ]

First, write all in prime bases:

9 = 3² → 9⁸ = (3²)⁸ = 3¹⁶
(x²)⁵ = x¹⁰
27 = 3³ → (27)⁴ = (3³)⁴ = 3¹²
(x³)² = x⁶

So expression becomes:

[3¹⁶ × x¹⁰] / [3¹² × x⁶] = 3^(16-12) × x^(10-6) = 3⁴ x⁴

Or (3x)⁴ — either is fine. But since it says “exponential form”, probably 3⁴ x⁴ is okay, or (3x)⁴. Both correct. I’ll write 3⁴ x⁴

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Now, compiling all final answers clearly.

Final Answer:

1.
a. 169
b. 125
c. 16
d. 121
e. -27
f. 1

2.
a. 300
b. 4000
c. 0
d. 27/64
e. 16/81

3.
a. (-5/7)^4
b. (-5)^3
c. x^5 a^2 b^3
d. (-2)^4 a^3

4.
a. 2^2 × 3^2
b. 3^3 × 5^2
c. 2^3 × 7^2
d. 2^5 × 3^3
e. 2 × 3^2 × 5^2
f. 2^3 × 3^2 × 5^2

5.
(i) 3^8
(ii) 7^3
(iii) 2^15
(iv) 10^4
(v) 4
(vi) 49/25

6.
(i) 2^5
(ii) 3^21
(iii) 2^8
(iv) x^3 y
(v) (2/5)^5
(vi) 3^4 x^4