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

## AP State Syllabus 7th Class Maths Solutions 11th Lesson Exponents InText Questions

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Question 1.

Write the following in exponential form, (values are rounded off) (Page No. 212)

i) Total surface area of the Earth is 510,000,000 square kilometers.

Solution:

51 × 10^{7} = 3× 17 × 10^{7}

ii) Population of Rajasthan is approximately 7,00,00,000.

Solution:

7 × 10^{7}

iii) The approximate age of the Earth is 4550 million years.

Solution:

4550 millions = 4550 × 10,00,000 (v 1 million =10 lakhs)

= 455 × 10^{7} = 91 × 5 × 10^{7} = 5 × 7 × 13 × 10^{7}

iv) 1000 km in meters.

Solution:

1 km = 1000 m

∴ 1000 km = 1000 × 1000 m = 10^{6}

Question 2.

Express (i) 48951 (ii) 89325 in expanded form using exponents. (Page No. 212)

Solution:

i) 48951 = (4 × 10000) + (8 × 1000) + (9 × 100) + (5 × 10) + (1 × 1)

= (4 × 104) + (8× 103) + (9 × 102) + (5 × 1.0) + (1 × 1)

ii) 89325 = (8 × 10000) + (9 × 1000) + (3 × 100) + (2 × 10) + (5 × 1)

= (8 × 10^{4}) + (9 × 10^{3}) + (3 × 10^{2}) + (2 × 10) + (5 × 1)

Question 3.

Is 3^{2} equal to 2^{3} ? Justify. (Page No. 213)

Solution:

3^{2} ≠ 2^{3}

Since 3^{2} = 3 × 3 = 9 and 2^{3} = 8

∴ 3^{2} ≠ 2^{3}

Question 4.

Write the following numbers in exponential form. Also state the

a) base b) exponent and c) how it is read.

i) 32 ii) 64 iii) 256 iv) 243 v) 49 (Page No. 213)

Solution:

i) 32 = 2 × 2 × 2 × 2 × 2 = 2^{5}

Base = 2; exponent = 5; read as 2 raised to the power 5.

ii) 64 = 2 × 2× 2 × 2 × 2 × 2 = 2^{6}

Base = 2; exponent = 6 and we read it as 2 raised to the power 6.

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

Base = 2, exponent = 8 and we read it as 2 raised to the power 8.

iv) 243 = 3 × 3 × 3 × 3 × 3 = 3^{5}

Base = 3; exponent = 5 and we read it as 3 raised to the power 5.

v) 49 = 7 × 7 = 7^{2}

= 7 is the base ; exponent = 2.

Question 5.

Write the expanded form of the following. (Page No. 213)

i) p^{7} ii) l^{4} iii) s^{9} iv) d^{6} v) z^{5}

Solution:

i) p^{7} = p × p × p × p × p × p × p

ii) l^{4} = l × l × l × l

iii) s^{9} = s × s × s × s × s × s × s × s × s

iv) d^{6} = d × d × d × d × d × d

v) z^{5} = z × z × z × z × z

Question 6.

Write the following in exponential form. (Page No. 213)

i) a × a × a × ………………….l’ times

ii) 5 × 5 × 5 × 5 × ……………..’n’ times

iii) q × q × q × q × q ………………….15 times

iv) r × r × r × ………………….’b’ times

Solution:

i) a × a × a × ………………….’l’ times = a^{l}

ii) 5 × 5 × 5 × 5 × ……………..’n’ times = 5^{n}

iii) q × q × q × q × q …………….15 times = q^{15}

iv) r × r × r × ……………..’b’ times = r^{b}

Do This

Question 1.

Find the values of 2^{4}, 2^{3} and 2^{7} and verify whether 2^{4} × 2^{3} = 2^{7}. (Page No. 215)

2^{4} = 2 × 2 × 2 ×2 = 16;

2^{3} = 2 × 2 x 2 = 8

2^{7} = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128

2^{4} × 2^{3} = 16 × 8 = 128 = 2^{7}

2^{4} × 2^{3} = 2^{7}

Question 2.

Find the values of 5^{2}, 5^{3} and 5^{5} and verify whether 5^{2} × 5^{3} = 5^{5}. (Page No. 215)

Solution:

5^{2} = 5 × 5 = 25;

5^{3} = 5 × 5 × 5 = 125 and 5^{5} = 5 × 5 × 5 × 5 × 5 = 3125

Now 52 × 53 = 25 × 125 = 3125 = 55

∴ 5^{2} × 5^{3} = 5^{5}

Question 3.

Simplify the following using the formula a^{m} × a^{n} = a^{m + n }(Page No. 216)

i) 3^{11} × 3^{9} ii) p^{5} × p^{8}

Solution:

i) 3^{11} × 3^{9} = 3^{11+9} = 3^{20}

ii) p^{5} × p^{8} = p^{5+8} = p^{13}

Question 4.

Find the appropriate number in place of the symbol’?’in the following. (Page No. 216)

Let ‘k’ be any non-zero integer.

i) k^{3} × k^{4} = k^{?}

Solution:

i) k^{3} × k^{4} = k^{?}

as k^{3} × k^{4} = k^{3+4} = k^{7} the value of ‘?’ = 7

ii) k^{15} × k^{?} = k^{31}

as k^{15} × k^{?} = k^{15+?}

but k^{15 + ?} = k^{31}

Since bases are equal we equate the exponents

∴ 15 + ? = 31

(i.e„) ? = 31 – 15 = 16

Question 5.

Compute 3^{6}, cube of 3^{2} and verify whether (3^{2})^{3} = 3^{6}. (Page No. 216)

Solution:

3^{6} = 3 × 3 × 3 × 3 × 3 × 3 = 729.

cube of 3^{2} = (3^{2})^{3} = 9^{3} = 9 × 9 × 9 = 729

Now (3^{2})^{3} = 3^{2} × 3^{2} × 3^{2} = 9 × 9 × 9 = 729

3^{6} = 3 × 3 × 3 × 3 × 3 × 3 = 9 × 9 × 9

(3^{2})^{3} = 3^{6}

Question 6.

Simplify the following using the law a^{m} × b^{m} = (ab)^{m }(Page No. 218)

i) (2 × 3)^{4}

ii) x^{p} × y^{p}

iii) a^{8} × b^{8}

iv) (5 × 4)^{11}

Solultion:

i) (2 × 3)^{4} = 2^{4} × 3 ^{4} = (2 × 2 × 2 × 2) × (3 × 3 × 3 ×3) = 16 × 81 = 1296

ii) x^{p} × y^{p} = (x . y)^{p}

iii) a^{8} × b^{8} = (a.b)^{8}

iv) (5 × 4)^{11} = 5^{11} × 4^{11} = 5^{11} × (2 × 2)^{11}

= 5^{11} × 2^{11} × 2^{11} = (5 × 2)^{11} × 2^{11} = 10^{11} × 2^{11}

Question 7.

Write the following, by using \(\mathbf{a}^{-n}=\frac{1}{\mathbf{a}^{n}}\) with positive exponents. (Page No. 219)

Solution:

Question 8.

Simplify and write in the form of a^{m-n} or \(\frac{1}{\mathbf{a}^{\mathbf{n}-\mathbf{m}}}\)

i) \(\frac{13^{8}}{13^{5}}\)

ii) \(\frac{3^{4}}{3^{14}}\)

Solution:

Question 9.

Fill the appropriate number in the box. (Page No. 222)

Solution:

Question 10.

Complete the following (Page No. 223)

Solution:

Question 11.

Write in expanded form. (Page No. 224)

i) a^{-5}

ii) (-a)^{4}

iii) (-7)^{-5}

iv) (-a)^{m}

Solution:

Question 12.

Write in exponential form. (Page No. 224)

i) (-3) × (-3) × (-3)

ii) (-b) × (-b) × (-b) × (-b)

iii) \(\left(\frac{1}{-2}\right) \times\left(\frac{1}{-2}\right) \times\left(\frac{1}{-2}\right)\) ………………….’m’ times

Solution:

i) (-3) × (-3) × (-3) = (-3)^{3}

ii) (-b) × (-b) × (-b) × (-b) = (-b)^{4}

iii) \(\left(\frac{1}{-2}\right) \times\left(\frac{1}{-2}\right) \times\left(\frac{1}{-2}\right)\) ………………….’m’ times = \(\left(-\frac{1}{2}\right)^{m}\) or (-2)^{-m}

Do This

Question 1.

Write the following in exponential form using prime factorization. (Page – 214)

i) 2500 ii) 1296 iii) 8000 iv)6300

Solution:

i) 2500 = 2 × 1250 = 2 × 2 × 625

= (2 × 2) × 5 × 125

= (2 × 2) × 5 × 5 × 25

= (2 × 2) × (5 × 5 × 5 × 5)

= 2^{2} × 5^{4}

ii) 1296 = 2 × 648 = 2 × 2 × 324 = 2 × 2 × 2 × 162

= (2 × 2 × 2 × 2) × 81

= (2 × 2 × 2 × 2 ) × 3 × 27

= (2 × 2 × 2 × 2 ) × 3 × 3 × 9

= (2 × 2 × 2 × 2 ) × ( 3 × 3 × 3 × 3 )

= 2^{4} × 3^{4}

iii) 8000 = 2 × 4000 = 2 × 2 × 2000 = 2 × 2 × 2 × 1000

= 2 × 2 × 2 × 2 × 500

= 2 × 2 × 2 × 2 × 2 × 250

= (2 × 2 × 2 × 2 × 2 × 2) × 125

= (2 × 2 × 2 × 2 × 2 × 2) × 5 × 25

= (2 × 2 × 2 × 2 × 2 × 2) × ( 5 × 5 × 5)

= (2^{6} × 5^{3})

iv) 6300 = 2 × 3150 = 2 × 2 × 1575

= (2 × 2) × 3 × 525

= 2 × 2 × 3 × 3 × 175

= (2 × 2) × (3 × 3) × 5 × 35

= (2 × 2) × (3 × 3) × (5 × 5) × 7

= 2^{2} × 3^{2} × 5^{2} × 7