# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

## AP Board Class 7 Maths Chapter 11 Exponents Ex 2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 11 Exponents Ex 2 Book Answers AP Board Class 7 Maths Chapter 11 Exponents Ex 2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 11 Exponents Ex 2 Book Answers

## Andhra Pradesh State Board Class 7th Maths Chapter 11 Exponents Ex 2 Books Solutions

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## AP Board Class 7th Maths Chapter 11 Exponents Ex 2 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 7th Maths Chapter 11 Exponents Ex 2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

Question 1.
Simplify the following using laws of exponents.
(i) 210 × 24
(ii) (32) × (32)4
(iii) 5752
(iv) 92 × 918 × 910
(v) (35)4×(35)3×(35)8
(vi) (-3)3 × (-3)10 × (-3)7
(vii) 3(2)2
(viii) 24 × 34
(ix) 24a × 25a
(x) (102)3
(xi) [(−56)2]5
(xii) 23a+7 × 27a+3
(xiii) (23)5
(xiv) (-3)3 × (-5)3
(xv) (−4)6(−4)3
(xvi) (23)5
(xvii) (−6)5(−6)9
(xviii) (-7)7 × (-7)8
(xix) (-64)4
(xx) ax × ay × a x
Solution:
(i) 210 × 24 = 210+4 = 214
[∵ am × an = am+n]

(ii) (32) × (32)4 = (32)1+4 = (32)5
= 32×5
= 310
[∵ am × an = am+n]
[∵ (am)= (a)mn]

(iii) 5752 = 57 – 2 = 55
= 5 × 5 × 5 × 5 × 5 = 55
[∵ 𝑎𝑚𝑎𝑛 = am-n, m > n]

(iv) 92 × 918 × 910 = 92+18+10 = 930
[∵ am × an = am+n]

(v) (35)4×(35)3×(35)8 = (35)4+3+8 = (35)15
[∵ am × an = am+n]

(vi) (-3)3 × (-3)10 × (-3)7 = (-3)3 + 10 + 7 = (-3)20
[∵ am × an = am+n]

(vii) 3(2)2 = 32×2 = 34
[∵ (am)n = amn])

(viii) 24 × 34 = (2 × 3 )4 = 64
[∵ am × bm = (ab)m]

(ix) 24a × 25a = 24a+5a = 29a
[∵ am × an = am+n]

(x) (102)3 = 102×3 = 106
[∵ (am)n = am×n ]

(xi) [(−56)2]5
[(−56)2]5=[−56]2×5 = (−56)10=(56)10 [∵ 10 even number]
[∵ (am)n = an]

(xii) 23a+7 × 27a+3
23a+7+7a+3 = = 210a+10 = 210(a+1)
[∵ am × an = (a)m+n]

(xiii) (23)5 = 2535
[∵ (𝑎𝑏)𝑚=𝑎𝑚𝑏𝑚 ]

(xiv) (-3)3 × (-5)3 = [(-3) × (-5)]3 = (15)3
[∵ am × bm = (ab)m]

(xv) (−4)6(−4)3 = (-4)3
[∵ 𝑎𝑚𝑎𝑛 = am-n, m > n]

(xvi) (23)5 = 1915−7 = 198 [∵ 𝑎𝑚𝑎𝑛=1𝑎𝑛−𝑚, n > m]

(xvii) (−6)5(−6)9 = 1(−6)9−5 [∵ 𝑎𝑚𝑎𝑛=1𝑎𝑛−𝑚, n > m]
1(−6)4=164 [∵4 is even number ]

(xviii) (-7)7 × (-7)8 = (-7)7+8 [∵ am × an = a m+n]
= (-7)15 = -(7)15 [∵ 15 is odd number ]

(xix) (-64)4 [∵ (am)n = amn]
= (-6)4×4 = (-6)16 = 616
[∵ 16 is even number ]

(xx) ax × ay × a x = ax+y+z
[am × an × ap = am+n+p]

Question 2.
By what number should 3 be multiplied so that the product is 729’?
Solution:
Given number = 3-4
Given product = 729 [∵ 36 = 729]
Let the number to be multiplied be x then
⇒ (3-4) . (x) = 36 ⇒ 𝑥34 =36
[∵ am × nn = am+n]
⇒ x = 36 × 34 = 310

Question 3.
1f 56 × 52x = 510 then find x.
Solution:
Given that 56 × 52x = 510
⇒ 56+2x = 510 [∵ am × an = am+n]
Since bases are equal, we equate the exponents
6 + 2x = 10
2x = 10 – 6 = 4
x = 42 = 2

Question 4.
Evaluate 20 + 30
Solution:
20 + 30 = 1 + 1 = 2 [∵ a0 = 1]

Question 5.
Simplify (𝑥𝑎𝑥𝑏)𝑎×(𝑥𝑏𝑥𝑎)𝑎×(𝑥𝑎𝑥𝑎)𝑏
Solution:
Given (𝑥axb)a×(xbxa)a×(xaxa)b=(xaxb×xbxa)a×(1)b
[∵ am × bm = (ab)m]
= 1a x 1b = 1a+b = 1

Question 6.
(i) 100 × 1011 = 1013
(ii) 32 × 43 = 125
(iii)) 5° = (100000)°
(iv) 43 = 82
(v) 23 > 32
(vi) (-2)4 > (-3)4
(vii) (-2)5 > (-3)5
Solution:
(i) 100 × 1011 = 1013 – True
as 100 × 1011 = 102 × 1011 = 102+11 = 1013

(ii) 32 × 43 = 125 – False
as 32 × 43 ≠ 125

(iii)) 5° = (100000)° – True as 5° = 1 and 100000° = 1

(iv) 43 = 82 – True, as 43 = 4 × 4 × 4 = 64 and 82 = 8 × 8 = 64

(v) 23 > 32 – False as 23 = 8 and 32 = 9 and 8 < 9

(vi) (-2)4 > (-3)4 – False, as (-2)4 = (-2) × (-2) × (-2) × (-2) = 16
(-3)4 = (-3) × (-3) × (-3) × (-3) = 81

(vii) (-2)5 > (-3)5 – True, as (-2)5 = (-2) × (-2) × (-2) × (-2) × (-2) = -32
(-3)5 = (-3) × (-3) × (-3) × (-3) × (-3) = -243

## Andhra Pradesh Board Class 7th Maths Chapter 11 Exponents Ex 2 Textbooks for Exam Preparations

Andhra Pradesh Board Class 7th Maths Chapter 11 Exponents Ex 2 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 7th Maths Chapter 11 Exponents Ex 2 exam preparation. The AP Board STD 7th Maths Chapter 11 Exponents Ex 2 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 7th Maths Chapter 11 Exponents Ex 2 Books State Board syllabus with maximum efficiency.

## FAQs Regarding Andhra Pradesh Board Class 7th Maths Chapter 11 Exponents Ex 2 Textbook Solutions

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