MP Board Class 7th Maths Solutions Chapter 13 Exponents and Powers Ex 13.2

MP Board Class 7th Maths Solutions Chapter 13 Exponents and Powers Ex 13.2

Question 1.
Using laws of exponents, simplify and write the answer in exponential form:
(i) 3² × 3⁴ × 3⁸
(ii) 6¹⁵ ÷ 6¹⁰
(iii) a³ × a²
(iv) 7^x × 7²
(v) (5²)³ 4 5³
(vi) 2⁵ × 5⁵
(vii) a⁴ × b⁴
(viii) (3⁴)³
(ix) (2²⁰ ÷ 2¹⁵) × 2³
(x) (8^t ÷ 8²)
Solution:
(i) 3² × 3⁴ × 3⁸ = (3)^{2 + 4 + 8} = 3¹⁴ [∵ a^m × aⁿ = a^m+n]
(ii) 6¹⁵ ÷ 6¹⁰ = (6)^{15 – 10} = 6⁵ [∵ a^m ÷ aⁿ = a^{m – n}]
(iii) a³ × a² = a^{3 + 2} = a⁵[∵ a^m × aⁿ = a^m+n]
(iv) 7^x × 7² = 7^{x + 2}[∵ a^m × aⁿ = a^m+n]
(v) (5²)³ ÷ 5³ = 5^{2 × 3} ÷ 5³ [∵ (a^m)ⁿ = a^mn]
= 5⁶ ÷ 5³ = 5^{(6 – 3)} [∵ a^m ÷ aⁿ = a^{m – n}]
= 5³
(vi) 2⁵ × 5⁵ = (2 × 5)⁵      [∵ a^m × b^m = (a × b)^m]
= 10⁵
(vii) a⁴ × b⁴ = (ab)⁴
(viii) (3⁴)³ = 3^{4 × 3} = 3¹²       [∵ (a^m)ⁿ = a^mn]
(ix) (2²⁰ ÷ 2¹⁵) × 2³ = (2^{20 – 15}) × 2³     [∵ a^m ÷ aⁿ = a^{m – n}]
= 2⁵ × 2³ = 2^{5 + 3} [∵ a^m × aⁿ = a^m+n]
(x) 8^t ÷ 8² = 8^{(t – 2)}    [∵ a^m ÷ aⁿ = a^{m – n}]

Question 2.
Simplify and express each of the following in exponential form:

Solution:

(ii) [(5²)³ × 5⁴] ÷ 5⁷
= [5^{2 × 3} × 5⁴] ÷ 5⁷ [∵ (a^m)ⁿ = a^mn]
= [5⁶ × 5⁴] ÷ 5⁷ [∵ a^m × aⁿ = a^{m + n}]
= [5^{6 + 4}] ÷ 5⁷
= 5¹⁰ ÷ 5⁷
= 5^{10 – 7} [∵ a^m ÷ aⁿ = a^{m – n}]
= 5³

(iii) 25⁴ ÷ 5³ = (5 × 5)⁴ ÷ 5³
= (5²)⁴ ÷ 5³ = 5^{2 × 4} ÷ 5³ [∵ (a^m)ⁿ = a^mn]
= 5⁸ ÷ 5³ = 5^{8 – 3} [∵ a^m ÷ aⁿ = a^{m – n}]
= 5⁵

(vi) 2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3     [∵ a⁰ = 1]
(vii) 2⁰ × 3⁰ × 4⁰ = 1 × 1 × 1 = 1
(viii) (3⁰ + 2⁰) × 5⁰ = (1 + 1) × 1 = 2

Question 3.
Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
(ii) 2³ > 5²
(iii) 2³ × 3² = 6⁵
(iv) 3⁰ = (1000)⁰
Solution:
(i) L.H.S = 10 × 10¹¹
= 10^{1 + 11} = 10¹²
R.H.S = 100¹¹ = (10 × 10)¹¹ = (10⁰)¹¹ = 10^{2 × 11} = 10²²
⇒ L.H.S. ≠ R.H.S.
Hence, the given statement is false.

(ii) 2³ > 5²
L.H.S. = 2³ = 2 × 2 × 2 = 8
R.H.S. = 5³ = 5 × 5 = 25
⇒ L.H.S ≠ R.H.S
Hence, the given statement is false.

(iii) 2³ × 3² = 6⁵
L.H.S. = 2³ × 3² = 2 × 2 × 2 × 3 × 3 = 72
R.H.S. = 6⁵ = 6 × 6 × 6 × 6 × 6 = 7776
⇒ L.H.S. ≠ R.H.S.
Hence, the given statement is false.

(iv) 3⁰ = (1000)⁰
L.H.S. = 3⁰ = 1
R.H.S. = (1000)⁰ = 1
⇒ L.H.S. = R.H.S.
Hence, the given statement is true.

Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768
Solution:
(i) 108 × 192
= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)
= (2² × 3³) × (2⁶ × 3)
= 2²⁺⁶ × 3³⁺¹ = 2⁸ × 3⁴
(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3³ × 5
(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2) = 3⁶ × 2⁶
(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3

Question 5.
Simplify:

Solution:

MP Board Class 7th Maths Solutions