AP State Syllabus AP Board 7th Class Maths Solutions Chapter 11 Exponents Ex 1 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 11th Lesson Exponents Exercise 1

Question 1.

Write the base and the exponent in each case. Also, write the term in the expanded form.

(i) 34

(ii) (7x)^{2}

(iii) (5ab)^{3}

(iv) (4y)^{5}

Solution:

Question 2.

Write the exponential form of each expression.

(i) 7 × 7 × 7 × 7 × 7

(ii) 3 × 3 × 3 × 5 × 5 × 5 × 5

(iii) 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5

Solution:

(i) 7 × 7 × 7 × 7 × 7 = 7^{5}

(ii) 3 × 3 × 3 × 5 × 5 × 5 × 5 = 3^{3} × 5^{4}

(iii) 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 2^{3} × 3^{4} × 5^{3}

Question 3.

Express the following as the product of exponents through prime factorization.

(i) 288

(ii) 1250

(iii) 2250

(iv) 3600

(v) 2400

Solution:

(i) 288

1) 288 = 2 × 144

= 2 × 2 × 72

= 2 × 2 × 2 × 36

= 2 × 2 × 2 × 2 × 18

= 2 × 2 × 2 × 2 × 2 × 9

= (2 × 2 × 2 × 2× 2) × (3 × 3)

= 2^{5} × 3^{2}

(ii) 1250 = 2 × 625

= 2 × 5 × 125

= 2 × 5 × 5 × 25

= 2 × 5 × 5 × 5 × 5

= 2 × 5^{4}

(iii) 2250 = 2 × 1125

= 2 × 3 × 375

= 2 × 3 × 3 × 125

= 2 × 3 × 3 × 5 × 25

= 2 × 3 × 3 × 5 × 5 × 5

= 2^{1} × 3^{2} × 5^{3}

(iv) 3600 = 2 × 1800

= 2 × 2 × 900

=2 × 2 × 2 × 450

=2 × 2 × 2 × 2 × 225

= ( 2 × 2 × 2 × 2 ) × 3 × 75

= (2 × 2 × 2 × 2) × (3 × 3) × 25

= (2 × 2 × 2 × 2) × (3 × 3) × (5 × 5)

= 2^{4} × 3^{2} × 5^{2}

(v) 2400

2400 = 2 × 1200

= 2 × 2 × 600

= 2 × 2 × 2 × 300

= 2 × 2 × 2 × 2 × 150

= 2 × 2 × 2 × 2 × 2 × 75

= (2 × 2 × 2 × 2 × 2) × 3 × 25

= (2 × 2 × 2 × 2 × 2) × 3 × (5 × 5)

= 2^{5} × 3^{1}

Question 4.

Identify the greater number in each of the following pairs.

(i) 2^{3} or 3^{2}

(ii) 5^{3} or 3^{5}

(iii) 2^{8} or 8^{2}

Solution:

(i) 2^{3} or 3^{2} = 2^{3} = 2 × 2 × 2 = 8 and 3^{2} = 3 × 3 = 9

∴ 2^{3} < 3^{2} or 3^{2} > 2^{3}

(ii) 5^{3} or 3^{5} = 5^{3} = 5 × 5 × 5 = 125 and 3^{5} = 3 × 3 × 3 × 3 × 3 = 243

∴ 3^{5} > 5^{3}

(iii) 2^{8} or 8^{2} = 2^{8} = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

8^{2} = 8 × 8 = 64

∴ 2^{8} > 8^{2}

Question 5.

If a = 3, b = 2 find the value of

(i) a^{b} + b^{a}

(ii) a^{a} + b^{b}

(iii) (a + b)a^{b}

(iv)(a – b)^{a}

Solution:

(i) a^{b} + b^{a} = 3^{2} + 2^{3} = 3 × 3 + 2 × 2 × 2 = 9 + 8 =17

(ii) a^{a} + b^{b} = 3^{3} + 2^{2} = 3 × 3 × 3 + 2 × 2 = 27 + 4 = 31

(iii) (a + b)^{b} = (3 + 2)^{2} = 5^{2} = 5 × 5 = 25

(iv)(a – b)^{a} = (3 – 2)^{2}= 1^{2} = 1 × 1 = 1 .