AP Board 7th Class Maths Solutions Chapter 11 Exponents Ex 2

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 11 Exponents Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Exponents Exercise 2

Question 1.
Simplify the following using laws of exponents.
(i) 210 × 24
(ii) (32) × (32)4
(iii) $$\frac{5^{7}}{5^{2}}$$
(iv) 92 × 918 × 910
(v) $$\left(\frac{3}{5}\right)^{4} \times\left(\frac{3}{5}\right)^{3} \times\left(\frac{3}{5}\right)^{8}$$
(vi) (-3)3 × (-3)10 × (-3)7
(vii) 3(2)2
(viii) 24 × 34
(ix) 24a × 25a
(x) (102)3
(xi) $$\left[\left(\frac{-5}{6}\right)^{2}\right]^{5}$$
(xii) 23a+7 × 27a+3
(xiii) $$\left(\frac{2}{3}\right)^{5}$$
(xiv) (-3)3 × (-5)3
(xv) $$\frac{(-4)^{6}}{(-4)^{3}}$$
(xvi) $$\left(\frac{2}{3}\right)^{5}$$
(xvii) $$\frac{(-6)^{5}}{(-6)^{9}}$$
(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)n = (a)mn]

(iii) $$\frac{5^{7}}{5^{2}}$$ = 57 – 2 = 55
= 5 × 5 × 5 × 5 × 5 = 55
[∵ $$\frac{a^{m}}{a^{n}}$$ = am-n, m > n]

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

(v) $$\left(\frac{3}{5}\right)^{4} \times\left(\frac{3}{5}\right)^{3} \times\left(\frac{3}{5}\right)^{8}$$ = $$\left(\frac{3}{5}\right)^{4+3+8}$$ = $$\left(\frac{3}{5}\right)^{15}$$
[∵ 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) $$\left[\left(\frac{-5}{6}\right)^{2}\right]^{5}$$
$$\left[\left(\frac{-5}{6}\right)^{2}\right]^{5}=\left[\frac{-5}{6}\right]^{2 \times 5}$$ = $$\left(\frac{-5}{6}\right)^{10}=\left(\frac{5}{6}\right)^{10}$$ [∵ 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) $$\left(\frac{2}{3}\right)^{5}$$ = $$\frac{2^{5}}{3^{5}}$$
[∵ $$\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}$$ ]

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

(xv) $$\frac{(-4)^{6}}{(-4)^{3}}$$ = (-4)3
[∵ $$\frac{a^{m}}{a^{n}}$$ = am-n, m > n]

(xvi) $$\left(\frac{2}{3}\right)^{5}$$ = $$\frac{1}{9^{15-7}}$$ = $$\frac{1}{9^{8}}$$ [∵ $$\frac{a^{m}}{a^{n}}=\frac{1}{a^{n-m}}$$, n > m]

(xvii) $$\frac{(-6)^{5}}{(-6)^{9}}$$ = $$\frac{1}{(-6)^{9-5}}$$ [∵ $$\frac{a^{m}}{a^{n}}=\frac{1}{a^{n-m}}$$, n > m]
$$\frac{1}{(-6)^{4}}=\frac{1}{6^{4}}$$ [∵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 ⇒ $$\frac{x}{3^{4}}$$ =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 = $$\frac{4}{2}$$ = 2

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

Question 5.
Simplify $$\left(\frac{x^{a}}{x^{b}}\right)^{a} \times\left(\frac{x^{b}}{x^{a}}\right)^{a} \times\left(\frac{x^{a}}{x^{a}}\right)^{b}$$
Solution:
Given $$\left(\frac{x^{\mathrm{a}}}{\mathrm{x}^{\mathrm{b}}}\right)^{\mathrm{a}} \times\left(\frac{\mathrm{x}^{\mathrm{b}}}{\mathrm{x}^{\mathrm{a}}}\right)^{\mathrm{a}} \times\left(\frac{\mathrm{x}^{\mathrm{a}}}{\mathrm{x}^{\mathrm{a}}}\right)^{\mathrm{b}}=\left(\frac{\mathrm{x}^{\mathrm{a}}}{\mathrm{x}^{\mathrm{b}}} \times \frac{\mathrm{x}^{\mathrm{b}}}{\mathrm{x}^{\mathrm{a}}}\right)^{\mathrm{a}} \times(1)^{\mathrm{b}}$$
[∵ 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