## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 4 Exponents and Powers Ex 4.2

Question 1.

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

Solution:

Question 2.

Simplify and write the following in the exponential form:

(i) 5^{3} × 5^{7} × 5^{12}

(ii) a^{5} × a^{3} × a^{7}

(iii) (7^{12} × 7^{3}) ÷ 7^{4}

Solution:

(i) 5^{3} × 5^{7} × 5^{12} = 5^{3+7+12} = 5^{22}

{∵ a^{m} × a^{n} × a^{p} = a^{m+n+p}}

(ii) a^{5} × a^{3} × a^{7} = a^{5+3+7} = a^{15}

(iii) (7^{12}× 7^{3}) ÷ 7^{4} = 7^{12+3} ÷ 7^{4} = 7^{15} – 7^{4} = 7^{15-4} = 7^{11}

Question 3.

Simplify and write the following in the exponential form:

(i) (2^{2})^{100}

(ii) ((-7)^{6})^{5}

(iii) (3^{2})^{5} × (3^{4})^{7}

Solution:

Using (a^{m})^{n} = a^{mn}

(i) (2^{2})^{100} = 2^{2×100} = 2^{200} {(a^{m})^{n} = a^{mn}}

(ii) [(-7)^{6}]^{5} = (-7)^{6×5} = (-7)^{30}

(iii) (3^{2})^{5} × (3^{4})^{7} = 3^{2×5} × 3^{4×7} = 3^{10} × 3^{28} = 3^{10+28} = 3^{38}

Question 4.

Simplify and write in the exponential form:

Solution:

Question 5.

Simplify and write in the exponential form:

Solution:

Question 6.

Simplify and express each of the following in the exponential form:

Solution:

Question 7.

Express each of the following rational numbers in the exponential form:

Solution:

Question 8.

Simplify the following:

Solution:

Question 9.

Simplify the following:

Solution:

Question 10.

Simplify the following:

Solution:

Question 11.

Simplify:

Solution:

Question 12.

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

(i) 108 × 192

(ii) 729 × 64

(iii) 384 × 147

Solution:

Question 13.

Simplify and write the following in the exponential form:

(i) 3^{3} × 2^{2} + 2^{2} × 5^{0}

(ii) 9^{2} + 11^{2} – 2^{2} × 3 × 17^{0}

Solution:

Question 14.

(i) By what number should we multiply 3^{4} so that the product is 3^{7}?

(ii) By what number should be multiple (-6)^{-1} so that the product is 10^{-1}?

Solution:

(i) Product = 3^{7}

Required number = 3^{7} ÷ 3^{4} = 3^{7-4} = 3^{3} = 27

(ii) Product = 10^{-1}

Required number = 10^{-1} ÷ (-6)^{-1}

Question 15.

Solution: