Exponents and Powers Class 7 Extra Questions with Answers

These Class 7 Maths Extra Questions Chapter 11 Exponents and Powers will help students prepare well for the exams.

Class 7 Maths Chapter 11 Extra Questions Exponents and Powers

Class 7 Maths Exponents and Powers Extra Questions – 2 Marks

Question 1.
i) 2³
Solution:
2³ = 2 × 2 × 2 = 8

ii) 8³
Solution:
8³ = 8 × 8 × 8 = 512

Question 2.
i) (-1)³
Solution:
(-1)³ = – 1 × – 1 × – 1 = – 1

ii) 5⁴
Solution:
5⁴ = 5 × 5 × 5 × 5 = 25 × 25 = 625

Question 3.
i) 2⁶
Solution:
2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

ii) 6³
Solution:
6³ = 6 × 6 × 6 = 216

Question 4.
i) (2024)°
Solution:
(2024)° = 1

ii) 8²
Solution:
8² = 8 × 8 = 64

Question 5.
Write any two laws of exponents.
Solution:
1) a^m × aⁿ = a^m-n
2) \(\frac{a^m}{a^n}\) = a^m-n
3) (a × b)^m = a^m × b^m
4) \(\frac{a^m}{b^m}\) = \(\left(\frac{a}{b}\right)^m\)
5) (a^m)ⁿ = a^m×n
6) a° = 1

Question 6.
i) (2³)²
Solution:
(2³)² = 2^3×2 = 2⁶ = 64

ii) (3²)⁴
Solution:
(3²)⁴ = 3^2×4 = 3⁸

Question 7.
i) 64^0.5
Solution:
(64)^0.5 = (8)^0.5 = 8^2×0.5 = 8¹ = 8

ii) 81^3/4
Solution:
81^3/4 = \(\left(3^4\right)^{\frac{3}{4}}\) = \(3^{4 \times \frac{3}{4}}\) = 3³ = 27

Question 8.
i) 10³
Solution:
10³ = 10 × 10 × 10 = 1000

ii) 2⁷
Solution:
2⁷ = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128

Question 10.
i) 4³
Solution:
4³ = 4 × 4 × 4 = 64

ii) 3⁵
Solution:
3⁵ = 3 × 3 × 3 × 3 × 3 = 243

Exponents and Powers Extra Questions Class 7 – 3 Marks

Question 11.
Find m, if 5^m × 5^p = 5²⁰
Solution:
5^m × 5^p = 5²⁰
5^m+p = 5²⁰ (∵ a^m × aⁿ = a^m+n)
As the bases are equal, we can equate power
m + p = 20
m = 20 – p

Question 12.
Express 1024 in exponential form.
Solution:
1024

1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2¹⁰

Question 13.
Express in exponential form 125 × 64.
Solution:

Question 14.
Evaluate:
i) (-1)²⁰²³
Solution:
(-1)²⁰²³ = -1 (∵ (-1)^{old number} = -1)

ii) (-1)²⁰²⁴
Solution:
(-1)²⁰²⁴ = 1 (∵ (-1)^{even number} = 1)

iii) (-1)°
Solution:
(-1)° = 1 (∵ a° = 1)

Question 15.
Simplify : \(\frac{\left(2^5\right)^2 \times 7^2}{8^2}\)
Solution:

Question 16.
If \(\left(\frac{3}{2}\right)^x\) × \(\left(\frac{3}{2}\right)^{-3}\) = \(\left(\frac{3}{2}\right)^7\) find x.
Solution:
\(\left(\frac{3}{2}\right)^x\) × \(\left(\frac{3}{2}\right)^{-3}\) = \(\left(\frac{3}{2}\right)^7\)
\(\left(\frac{3}{2}\right)^x-3\) = \(\left(\frac{3}{2}\right)^{7}\) (∵ a^m-n × a = a^m-n)
As the bases are equal then we can equate powers.
x – 3 = 7 ⇒ x = 7 + 3 ⇒ x = 10

Question 17.
i) 25² ÷ 5²
Solution:
25² ÷ 5² = \(\frac{\left(5^2\right)^2}{5^2}\) = \(\frac{5^4}{5^2}\)
(∵ \(\frac{a^m}{a^n}\) = a^m-n)
= 5^4-2
= 5²

Question 18.
Simplify (3¹⁵ × 3¹⁶) ÷ 3⁸
Solution:
(3¹⁵ × 3¹⁶) ÷ 3⁸
= (3¹⁵⁺¹⁶) ÷ 3⁸ (∵ a^m × aⁿ = a^m+n)
= 3³¹ ÷ 3⁸ = \(\frac{3^{31}}{3^8}\)
= 3^{31 – 8}
(∵ \(\frac{a^m}{a^n}\) = a^m-n)
= 3²³

Question 19.
Simplify \(\frac{\left(\mathbf{a}^9 \times b^6\right) \times b^3}{a^2 b^2 c^3}\)
Solution:

Question 20.
Simplify
i) (2° + 16°) × 19°
Solution:
(2° + 16°) × 19° = (1 + 1) × 1 = 2 × 1 = 2

ii) (6° + 5°) × 4°
Solution:
(6° + 5°) × 4° = (1 – 1) × 1 = 0 × 1 = 0

Extra Questions of Exponents and Powers Class 7 – 5 Marks

Question 21.
If 2^r-3 × 2^7r+1 = 2¹⁶ find r.
Solution:
2^r-3 × 2^7r+1 = 2¹⁶
2^r-3+7r-1 = 2¹⁶ (∵ a^m × aⁿ = a^m+n)
2^8r-2 = 2¹⁶
8r – 2 = 16 ⇒ 8r = 16 + 2
8r = 18 ⇒ r = \(\frac { 18 }{ 8 }\) ⇒ r = \(\frac { 9 }{ 4 }\)

Question 22.
If (a^m)ⁿ = a^mn, then express m in terms of n.
Solution:
Given (a^m)ⁿ = a^mn ⇒ a^mn = a^mn
mn = mⁿ ⇒ n = \(\frac{m^n}{m^1}\)
n = m^n-1 ⇒ m = \((n)^{\frac{1}{n-1}}\)

Question 23.
If 5^x = 1000, then find

i) 5^x+2
Solution:
Given 5^x = 1000
5^x-2 = 5^x × 5² = 1000 × 25 = 25,000

ii) 5^x-2
Solution:
5^x-2 = 5^x × 5^-2 = \(\frac{5^x}{5^2}\) = \(\frac { 1000 }{ 25 }\) = 40

Question 24.
Find k so that
(-3)^k-1 × (-3)^-5 = (-3)⁸
Solution:
(-3)^k-1 × (-3)^-5 = (-3)⁸
(-3)^k-1-5 = (-3)⁸
(-3)^k-6 = (-3)⁸
k – 6 = 8 ⇒ k = 8 + 6 ⇒ k = 14

Question 25.
If 16 × 2^p+2 = 32, find p
Solution:
16 × 2^p+2 = 32
2⁴ × 2^p+2 = 2⁵
2^4+p+2 = 2⁵ ⇒ 2^6+p = 2⁵
6 + p = 5 ⇒ p = 5 – 6 ⇒ p = – 1

Question 26.
Evaluate \(\left(\frac{2}{5}\right)^3\) × \(\left(\frac{5}{2}\right)^4\) × \(\frac { 3 }{ 2 }\)
Solution:
\(\left(\frac{2}{5}\right)^3\) × \(\left(\frac{5}{2}\right)^4\) × \(\frac { 3 }{ 2 }\) = \(\frac{2^3}{5^3}\) × \(\frac{5^4}{2^4}\) × \(\frac{3}{2}\)
= 2^3-4-2 × 3 × 5^4-2 = 2⁵ × 3 × 5²

Question 27.
If 25^n-1 ÷ 5² = 5¹⁰ find n.
Solution:
25^n-1 ÷ 5² = 5¹⁰
(5²)^n-1 ÷ 5² = 5¹⁰
\(\frac{5^{2 n-2}}{5^2}\) = 5¹⁰ (∵ (a^m)ⁿ = a^mn)
5^2n-2-2 = 5¹⁰
(∵ \(\frac{a^m}{a^n}\) = a^m-n)
5^2n-2-2 = 5¹⁰ ⇒ 2n – 4 = 10
2n = 10 + 4 ⇒ 2n = 14 ⇒ n = \(\frac { 14 }{ 2 }\) ⇒ n = 7

Question 28.
Evaluate
\(\left(\frac{1}{2}\right)^0\) × \(\left(2^{-1}\right)^0\) × \(\left(\frac{1}{2}\right)^{-1}\) + \(\left(\frac{1}{2}\right)^{-2}\)
Solution:
\(\left(\frac{1}{2}\right)^0\) × \(\left(2^{-1}\right)^0\) × \(\left(\frac{1}{2}\right)^{-1}\) + \(\left(\frac{1}{2}\right)^{-2}\)
= 1 + 1 + 2 + 2² (∵ a° = 1; a^-n = \(\frac{1}{a^n}\))
= 1 + 3 + 4 = 8

Question 29.
Write the filling in usual form
i) 5.74 × 10⁵
Solution:
5.74 × 10⁵ = 574 × 10³ = 574000

ii) 9.235 × 10¹¹
Solution:
9.235 × 10¹¹ = 9235 × 10⁸ = 923500000000

Question 30.
Express in standard form:
i) 3,70,978
Solution:
3,70,978 = 3,70,978 × 10⁵

ii) 64,950,000
Solution:
64,950,000 = 6495 × 10⁴ = 6.495 × 10³ × 10⁴ = 6.495 × 10⁷