Well-designed AP 7th Class Maths Textbook Solutions Chapter 11 Exponents and Powers Exercise 11.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Exponents and Powers Class 7 Exercise 11.2 Solutions – 7th Class Maths 11.2 Exercise Solutions

Question 1.

Using laws of exponents, simplify and write the answer in exponential form:

i) 3^{2} × 3^{4} × 3^{8}

Solution:

3^{2} × 3^{4} × 3^{8} = 3^{2+4+8} = 3^{14}

(∵ a^{m} × a^{n} × a^{p} = a^{m+n+p}

ii) 6^{15} ÷ 6^{10}

Solution:

6^{15} ÷ 6^{10} = 6^{15 – 10} = 6^{5}

(∵ a^{m} ÷ a^{n} = a^{m-n})

iii) a^{3} × a^{2}

Solution:

a^{3} × a^{2} = a^{3-2} = a^{5}

(∵ a^{m} × a^{n} = a^{m-n})

iv) 7^{x} × 7^{2}

Solution:

7^{x} × 7^{2} = 7^{x+2}

(∵ a^{m} × a^{n} =a^{m+n})

= 7^{x+2}

v) (5^{2})^{3} ÷ 5^{3}

Solution:

(5^{2})^{3} ÷ 5^{3}

= \(\frac{\left(5^2\right)^3}{5^3}\) = \(\frac{5^6}{5^3}\) (∵ (a^{m}) = a^{m×n})

= 5^{6-3} = 5^{3} = 125 \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)

vi) 2^{5} × 5^{5}

Solution:

2^{5} × 5^{5} = (2 × 5)^{3} [∵ a^{m} × b^{m} = (a × b)^{m}]

=10^{5} = 10 × 10 × 10 × 10 × 10 = 100000

vii) a^{4} × b^{4}

Solution:

a^{4} × b^{1} = (a × b)^{4} = (ab)^{4}

viii) (3^{4})^{3}

Solution:

(3^{4})^{3} = 3^{4×3} = 3^{12}

[∵ (a^{m})^{n} = a^{mn}]

ix) (2^{20} ÷ 2^{15}) × 2^{3}

Solution:

(2^{20} ÷ 2^{15}) × 2^{3} = \(\left(\frac{2^{20}}{2^{15}}\right)\) × 2^{3}

= (2^{20-15}) × 2^{3} \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)

= 2^{5} × 2^{3}

= 2^{5+3} (∵ a^{m} × a^{n} = a^{m+n})

= 2^{8}

x) 8^{t} ÷ 8^{2}

Solution:

8^{t} ÷ 8^{2} = 8^{t-2}

(∵ a^{m} ÷ a^{n} = ^{m-n})

ix) (2^{20} ÷ 2^{15}) × 2^{3}

Solution:

(2^{20} ÷ 2^{15}) × 2^{3} = \(\left(\frac{2^{20}}{2^{15}}\right) \times 2^3\)

= (2^{20 – 15} × 2^{3} \(\left( \frac{a^m}{a^n}=a^{m-n}\right)\)

= 2^{5} × 2^{3}

= 2^{5+3} (∵ a^{m} × a^{n} = a^{m+n})

= 2^{8}

x) 8^{t} ÷ 8^{2}

Solution:

8^{t} ÷ 8^{2} = 8^{t-2} (∵ a^{m} ÷ a^{n} = a^{m-n})

Question 2.

Simplify and express each of the following in exponential form :

v) \(\frac{3^7}{3^4 \times 3^3}\)

Solution:

\(\frac{3^7}{3^4 \times 3^3}\) = \(\frac{3^7}{3^{4+3}}\) (∵ a^{m} × a^{n} = a^{m+n})

= \(\frac{3^7}{3^7}\) \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)

= 3^{7–7} = 3° (∵ a° = 1)

= 1

vi) 2° + 3° + 4°

Solution:

2° + 3° + 4° = 1 + 1 + 1 = 3 (∵ a° = 1)

viii) (3° + 2°) × 5°

Solution:

(3° + 2°) × 5° = (1 + 1) × 1 = 2 × 1 = 2 (∵ a° = 1)

ix) \(\frac{2^8 \times a^5}{4^3 \times a^3}\)

Solution:

\(\frac{2^8 \times a^5}{4^3 \times a^3}\) = \(\frac{2^8 \times a^5}{\left(2^2\right)^3 \times a^3}\) (∵ (a^{m})^{n} = a^{mn})

= \(\frac{2^8 \times \mathrm{a}^5}{2^6 \times \mathrm{a}^3}\)

= 2^{8–6} × a^{5–3} \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)

= 2^{2} × a^{2}

= (2 × a)^{2} [∵ a^{m} × b^{m} = (a × b)^{m}]

= (2a)^{2}

x) \(\left(\frac{a^5}{a^3}\right) \times a^8\)

Solution:

\(\left(\frac{a^5}{a^3}\right) \times a^8\) = a^{8} = (a^{5–3}) × a^{8}

\(\left(\frac{x^m}{x^n}=x^{m-n}\right)\)

= a^{2} × a^{8} = a^{2+8} = a^{10} (∵ X^{m} × X^{n} = X^{m+n})

xi) \(\frac{4^5 \times a^8 b^3}{4^5 \times a^5 b^2}\)

Solution:

\(\frac{4^5 \times a^8 b^3}{4^5 \times a^5 b^2}\)

= 4^{5–5} × a^{8–5} × b^{3–2} \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)

= 4^{0} × a^{3} × b^{1} = 1 × a^{3} × b = a^{3}b (∵ a° = 1)

xii) (2^{3} × 2)^{2}

Solution:

(2^{3} × 2)^{2} = (2^{3} × 2^{1})^{2} (∵ a^{m} × a^{n} = a^{m+n})

= (2^{3+1})^{2}

= (2^{4})^{2} = 2^{4×2} (∵ (a^{m})^{n} = a^{m×n})

= 2^{8}

Question 3.

Say true or false and justify your answer :

i) 10 × 10^{11} = 100^{11}

Solution:

10 × 10^{11} = 100^{11}

LHS 10^{1} × 10^{11} = 10^{1+11} =10^{12} (∵ a^{m} × a^{n} = a^{m+n})

RHS 100^{11}

10^{12} ≠ 100^{11}

∴ 10 × 10^{11} ≠ 100^{11}

∴ False

ii) 2^{3} > 5^{2}

Solution:

2^{3} > 5^{2}

2^{3} = 8; 5^{2} = 25

8 > 25 (False)

But 8 < 25

∴ False

iii) 2^{3} × 3^{2} = 6^{5}

Solution:

2^{3} × 3^{2} = 6^{5}

LHS 2^{3} × 3^{2} = 8 × 9 = 72

RHS 6^{5} = 6 × 6 × 6 × 6 × 6 = 7776

7776 > 72

∴ False

iv) 3° = (1000)°

Solution:

3° = (1000)°

LHS 3° = 1

RHS (1000)° = 1

∴ LHS = RHS

∴ True

Question 4.

Express each of the following as a product of prime factors only in exponential form :

i) 108 × 192

Solution:

108 × 192

108 = 2 × 2 × 3 × 3 × 3

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

108 × 192 = 2 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2^{8} × 3^{4}

ii) 270

Solution:

270

270 = 2 × 5 × 3 × 3 × 3

270 = 2 × 5 × 3^{3}

iii) 729 × 64

Solution:

729 × 64

729 = 3 × 3 × 3 × 3 × 3 × 3 ⇒ 729 = 3^{6}

64 = 2 × 2 × 2 × 2 × 2 × 2

64 = 2^{6}; 729 × 64 = 3^{6} × 2^{6} = (3 × 2)^{6} = 6^{6}

iv) 768

Solution:

768

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

768 = 2^{8} × 3

Question 5.

Simplify:

i) \(\frac{\left(2^5\right)^3 \times 7^3}{8^3 \times 7}\)

Solution:

\(\frac{\left(2^5\right)^3 \times 7^3}{8^3 \times 7}\) = \(\frac{2^{5 \times 2} \times 7^3}{\left(2^3\right)^3 \times 7^1}\) (∵(a^{m})^{n} – a^{mn})

= \(\frac{2^{15} \times 7^3}{2^9 \times 7^1}\) = 2^{10-4} × 7^{3-1}

\(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)

= 2^{1} × 7^{2} – 2 × 7^{2} = 2 × 49 = 98

ii) \(\frac{25 \times 5^2 \times t^2}{10^3 \times t^4}\)

Solution:

iii) \(\frac{3^5 \times 10^5 \times 25}{5^7 \times 6^5}\)

Solution: