Rd Sharma 2020 2021 for Class 7 Maths Chapter 6 - Exponents

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Rd Sharma 2020 2021 Solutions for Class 7 Maths Chapter 6 Exponents are provided here with simple step-by-step explanations. These solutions for Exponents are extremely popular among Class 7 students for Maths Exponents Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 2021 Book of Class 7 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 2021 Solutions. All Rd Sharma 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

Page No 6.12:

Question 1:

Find the value of each of the following:
(i) 13²
(ii) 7³
(iii) 3⁴

Answer:

We have
(i) 13² = 13 × 13 = 169
(ii) 7³ = 7 × 7 × 7 = 343
(iii) 3⁴ = 3 × 3 × 3 × 3 = 81

Page No 6.12:

Question 2:

Find the value of each of the following:
(i) (−7)²
(ii) (−3)⁴
(iii) (−5)⁵

Answer:

We know that if 'a' is natural number, then
(−a)^{even number} = Positive number
(−a)^{odd number}= Negative number

We have
(i) (−7)² = −7 × −7 = 49
(ii) (−3)⁴ = −3 × −3 × −3 × −3 = 81
(iii) (−5)⁵ = −5 × −5 × −5 × −5 × −5 = −3125

Page No 6.12:

Question 3:

Simplify:
(i) 3 × 10²
(ii) 2² × 5³
(ii) 3³ × 5²

Answer:

We have
(i) 3 × 10² = 3 × 100 = 300             [since 10² = 10 × 10 = 100]
(ii) 2² × 5³= 4 × 125 = 500            [since 2² = 2 × 2 = 4 and 5³ = 5 × 5 × 5 = 125]
(iii) 3³ × 5² = 27 × 25 = 675           [ since 3³ = 3 × 3 × 3 = 27 and 5² = 5 × 5 = 25]

Page No 6.12:

Question 4:

Simplify:
(i) 3² × 10⁴
(ii) 2⁴ × 3²
(ii) 5² × 3⁴

Answer:

We have
(i) 3² × 10⁴ = 9 × 10000 = 90000       [since 3² = 3 × 3 = 9 and 10⁴ = 10 × 10 × 10 × 10 = 10000]
(ii) 2⁴ × 3² = 16 × 9 = 144                  [since 2⁴ = 2 × 2 × 2 × 2 = 16 and 3² = 3 × 3 = 9]
(iii) 5² × 3⁴ = 25 × 81 = 2025              [since 5² = 5 × 5 = 25 and 3⁴ = 3 × 3 × 3 × 3 = 81]

Page No 6.12:

Question 5:

Simplify:
(i) (−2) × (−3)³
(ii) (−3)² × (−5)³
(iii) (−2)⁵ × (−10)²

Answer:

We know that if 'a' is natural number, then
(−a)^{even number} = Positive number
(−a)^{odd number} = Negative number

We have

(i) (−2) × (−3)³ = ( −2 )(−27) = 54               [since (−3)³ = −3 ×−3 × − 3 = −27]
(ii) (−3)² × ( −5)³= 9 (−125) = −1125        [ since (−3)² = −3 ×− 3 = 9 and (−5 )³ = −5 ×−5 × − 5 = −125]
(iii) ( −2)⁵× (−10)² = −32 × 100 = −3200    [ since (−2)⁵= −2 ×−2 × −2 ×−2 ×−2 = −32 and (−10)² = −10 ×− 10 = 100]

Page No 6.12:

Question 6:

Simplify:
(i) ${(\frac{3}{4})}^{2}$
(ii) ${(\frac{- 2}{3})}^{4}$
(iii) ${(\frac{- 4}{5})}^{5}$

Answer:

We have

(i) ${(\frac{3}{4})}^{2} = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$
(ii) ${(\frac{- 2}{3})}^{4} = \frac{- 2}{3} \times \frac{- 2}{3} \times \frac{- 2}{3} \times \frac{- 2}{3} = \frac{16}{81}$
(iii) ${(\frac{- 4}{5})}^{5} = \frac{- 4}{5} \times \frac{- 4}{5} \times \frac{- 4}{5} \times \frac{- 4}{5} \times \frac{- 4}{5} = \frac{- 1024}{3125}$

Page No 6.12:

Question 7:

Identify the greater number in each of the following:
(i) 2⁵ or 5²
(ii) 3⁴ or 4³
(iii) 3⁵ or 5³

Answer:

We have
(i) 2⁵ = 2 × 2 × 2 × 2 × 2 = 32 and 5² = 5 × 5 = 25
     Therefore, 32 > 25.
     Thus, 2⁵ > 5².

(ii) 3⁴ = 3 × 3 × 3 × 3 = 81 and 4³= 4 × 4 × 4 = 64
      Therefore, 81 > 64.
      Thus, 3⁴ > 4³.

(iii) 3⁵ = 3 × 3 × 3 × 3 × 3 = 243 and 5³ = 5 × 5 × 5 = 125
       Therefore, 243 > 125.
       Thus, 3⁵ > 5³.

Page No 6.12:

Question 8:

Express each of the following in exponential form:
(i) (−5) × (−5) × (−5)
(ii) $\frac{- 5}{7} \times \frac{- 5}{7} \times \frac{- 5}{7} \times \frac{- 5}{7}$
(iii) $\frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} \times \frac{4}{3}$

Answer:

We have
(i) (−5) × (−5) × (−5) = ( −5)³

(ii) $\frac{- 5}{7} \times \frac{- 5}{7} \times \frac{- 5}{7} \times \frac{- 5}{7} = {(\frac{- 5}{7})}^{4}$

(iii) $\frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} \times \frac{4}{3} = {(\frac{4}{3})}^{5}$

Page No 6.12:

Question 9:

Express each of the following in exponential form:
(i) x × x × x × x × a × a × b × b × b
(ii) (−2) × (−2) × (−2) × (−2) × a × a × a
(iii) $(\frac{- 2}{3}) \times (\frac{- 2}{3}) \times x \times x \times x$

Answer:

We have
(i) $x \times x \times x \times x \times a \times a \times b \times b \times b = x^{4} a^{2} b^{3}$
(ii) $(- 2) \times (- 2) \times (- 2) \times (- 2) \times a \times a \times a = (- 2)^{4} \times a^{3}$
(iii) $(\frac{- 2}{3}) \times (\frac{- 2}{3}) \times x \times x \times x = {(\frac{- 2}{3})}^{2} \times x^{3}$

Page No 6.12:

Question 10:

Express each of the following numbers in exponential form:
(i) 512
(ii) 625
(iii) 729

Answer:

We have
(i) Prime factorisation of 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁹

(ii) Prime factorisation of 625 = 5 x 5 x 5 x 5 = 5⁴

(iii) Prime factorisation of 729 = 3 x 3 x 3 x 3 x 3 x 3 = 3⁶

Page No 6.12:

Question 11:

Express each of the following numbers as a product of powers of their prime factors:
(i) 36
(ii) 675
(iii) 392

Answer:

We have
(i) Prime factorisation of 36 = 2 x 2 x 3 x 3 = 2² x 3²

(ii) Prime factorisation of 675 = 3 x 3 x 3 x 5 x 5 = 3³ x 5²

(iii) Prime factorisation of 392 = 2 x 2 x 2 x 7 x 7 = 2³ x 7²

Page No 6.13:

Question 12:

Express each of the following numbers as a product of powers of their prime factors:
(i) 450
(ii) 2800
(iii) 24000

Answer:

We have
(i) Prime factorisation of 450 = 2 x 3 x 3 x 5 x 5 = 2 x 3² x 5²

(ii) Prime factorisation of 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7 = 2⁴ x 5² x 7

(iii) Prime factorisation of 24000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5 = 2⁶ x 3 x 5³

Page No 6.13:

Question 13:

Express each of the following as a rational number of the form $\frac{p}{q}$ :
(i) ${(\frac{3}{7})}^{2}$
(ii) ${(\frac{7}{9})}^{3}$
(iii) ${(\frac{- 2}{3})}^{4}$

Answer:

We have

(i) ${(\frac{3}{7})}^{2} = \frac{3}{7} \times \frac{3}{7} = \frac{9}{49}$
(ii) ${(\frac{7}{9})}^{3} = \frac{7}{9} \times \frac{7}{9} \times \frac{7}{9} = \frac{343}{729}$

(iii) ${(\frac{- 2}{3})}^{4} = \frac{- 2}{3} \times \frac{- 2}{3} \times \frac{- 2}{3} \times \frac{- 2}{3} = \frac{16}{81}$

Page No 6.13:

Question 14:

Express each of the following rational numbers in power notation:
(i) $\frac{49}{64}$
(ii) $- \frac{64}{125}$
(iii) $- \frac{1}{216}$

Answer:

We have
(i) $\frac{49}{64} = \frac{7}{8} \times \frac{7}{8} = \frac{(7)^{2}}{(8)^{2}} = {(\frac{7}{8})}^{2}$
(ii) $- \frac{64}{125} = - \frac{4}{5} \times - \frac{4}{5} \times - \frac{4}{5} = - \frac{(4)^{3}}{(5)^{3}} = {(- \frac{4}{5})}^{3}$
(iii) $- \frac{1}{216} = \frac{- 1}{6} \times \frac{- 1}{6} \times \frac{- 1}{6} = \frac{(- 1)^{3}}{(6)^{3}} = {(\frac{- 1}{6})}^{3}$

Page No 6.13:

Question 15:

Find the value of each of the following:
(i) ${(\frac{- 1}{2})}^{2} \times 2^{3} \times {(\frac{3}{4})}^{2}$
(ii) ${(\frac{- 3}{5})}^{4} \times {(\frac{4}{9})}^{4} \times {(\frac{- 15}{18})}^{2}$

Answer:

We have

(i)
${(\frac{- 1}{2})}^{2} \times 2^{3} \times {(\frac{3}{4})}^{2} = \frac{1}{4} \times 8 \times \frac{9}{16} = \frac{1}{4} \times \frac{9}{2} = \frac{9}{8} [Since {(\frac{- 1}{2})}^{2} = \frac{1}{4}, 2^{3} = 8 and {(\frac{3}{4})}^{2} = \frac{9}{16}]$

(ii)
${(\frac{- 3}{5})}^{4} \times {(\frac{4}{9})}^{4} \times {(\frac{- 15}{18})}^{2} = \frac{(- 3)^{4}}{5^{4}} \times \frac{4^{4}}{9^{4}} \times {(\frac{- 3 \times 5}{2 \times 9})}^{2} = \frac{(- 3)^{4}}{5^{4}} \times \frac{4^{4}}{9^{2} \times 9^{2}} \times {(\frac{- 3 \times 5}{2 \times 9})}^{2} = \frac{81}{5^{4}} \times \frac{4^{4}}{81 \times 9^{2}} \times \frac{(- 3)^{2} \times 5^{2}}{2^{2} \times 9^{2}} = \frac{1}{5^{4}} \times \frac{4^{4}}{9^{2}} \times \frac{(- 3)^{2} \times 5^{2}}{2^{2} \times 9^{2}} = \frac{1}{5^{4}} \times \frac{4^{4}}{9^{2}} \times \frac{9 \times 5^{2}}{4 \times 9^{2}} = \frac{1}{5^{2}} \times \frac{4^{3}}{9^{2}} \times \frac{1}{9} = \frac{1 \times 4^{3}}{5^{2} \times 9^{3}} = \frac{64}{25 \times 729} = \frac{64}{18225} [Since 4^{3} = 64, 5^{2} = 25 and 9^{3} = 729]$

Page No 6.13:

Question 16:

If a = 2 and b = 3, then find the values of each of the following:
(i) (a + b)^a
(ii) (ab)^b
(iii) ${(\frac{b}{a})}^{b}$
(iv) ${(\frac{a}{b} + \frac{b}{a})}^{a}$

Answer:

We have a = 2 and b = 3.
Thus,
(i) (a + b)^a = (2 + 3)² = (5)² = 25

(ii) (ab)^b = (2 x 3 )³ = (6)³ = 216

(iii) ${(\frac{b}{a})}^{b} = {(\frac{3}{2})}^{3} = \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} = \frac{27}{8}$

(iv) ${(\frac{a}{b} + \frac{b}{a})}^{a} = {(\frac{2}{3} + \frac{3}{2})}^{2} = {(\frac{4 + 9}{6})}^{2} = {(\frac{13}{6})}^{2} = \frac{169}{36}$

Page No 6.28:

Question 1:

Using laws of exponents, simplify and write the answer in exponential form:
(i) 2³ × 2⁴ × 2⁵
(ii) 5¹² ÷ 5³
(iii) (7²)³
(iv) (3²)⁵ ÷ 3⁴
(v) 3⁷ × 2⁷
(vi) (5²¹ ÷ 5¹³) × 5⁷

Answer:

We have

(i) 2³ x 2⁴ x 2⁵ = 2^{(3 + 4 + 5)} = 2¹²             [since a^m + aⁿ+ a^p = a^(m+n+p)]

(ii) 5¹² ÷ 5³ = $\frac{5^{12}}{5^{3}}$ = 5^{12 - 3} = 5⁹                [ since a^m÷ aⁿ= a^m-n ]
(iii) (7²)³ = 7⁶[since (a^m)ⁿ = a^mn ]

(iv)(3²)⁵ ÷ 3⁴ = 3¹⁰ ÷ 3⁴      [since (a^m)ⁿ = a^mn ]
                       = 3^{(10 - 4)} = 3⁶                      [since a^m÷ aⁿ= a^m-n ]

(v) 3⁷ × 2⁷ = (3 x 2)⁷ = 6⁷                          [since a^m x b^m = (a x b)^m ]

(vi) (5²¹ ÷ 5¹³) x 5⁷ = 5^{(21 -13)} x 5⁷         [since a^m÷ aⁿ= a^m-n ]
                                   = 5⁸ x 5⁷                  [since a^m x bⁿ =a^{(m +n)}]
                                 = 5⁽⁸⁺⁷⁾
                                   = 5¹⁵

Page No 6.28:

Question 2:

Simplify and express each of the following in exponential form:
(i) ${(2^{3})^{4} \times 2^{8}} \div 2^{12}$
(ii) (8² × 8⁴) ÷ 8³
(iii) $(\frac{5^{7}}{5^{2}}) \times 5^{3}$
(iv) $\frac{5^{4} \times x^{10} y^{5}}{5^{4} \times x^{7} y^{4}}$

Answer:

We have
(i) {(2³)⁴ x 2⁸} ÷ 2¹²
    = {2¹² x 2⁸} ÷ 2¹²
    = 2^{(12 + 8)} ÷ 2¹²   = 2²⁰ ÷ 2¹²= 2^{(20 - 12)}= 2⁸

(ii) (8² x 8⁴) ÷ 8³
= 8^{(2 + 4)} ÷ 8³
    = 8⁶ ÷ 8³
   = 8^(6-3) = 8³ = (2³)³ = 2⁹

(iii) $(\frac{5^{7}}{5^{2}})$ x 5³ = 5^(7-2) x 5³
                        = 5⁵ x 5³
                        = 5^{(5 + 3 )} = 5⁸

(iv) $\frac{5^{4} \times x^{10} y^{5}}{5^{4} \times x^{7} y^{4}}$ = $5^{(4 - 4)} \times x^{(10 - 7)} \times y^{(5 - 4)}$
                         = $5^{0} \times x^{3} \times y$                [since 5⁰ = 1]
                         = $1 \times x^{3} y = x^{3} y$

Page No 6.28:

Question 3:

Simplify and express each of the following in exponential form:
(i) ${(3^{2})^{3} \times 2^{6}} \times 5^{6}$
(ii) ${(\frac{x}{y})}^{12} \times y^{24} \times (2^{3})^{4}$
(iii) ${(\frac{5}{2})}^{6} \times {(\frac{5}{2})}^{2}$
(iv) ${(\frac{2}{3})}^{5} \times {(\frac{3}{5})}^{5}$

Answer:

We have
(i)   {(3²)³ x 2⁶} x 5⁶
    = {3⁶x 2⁶} x 5⁶             [since (a^m)ⁿ = a^mn]
    = 6⁶ x 5⁶                        [since a^m x b^m= (a x b)^m ]
    = 30⁶

(ii)
${(\frac{x}{y})}^{12} \times y^{24} \times (2^{3})^{4}$
$= \frac{x^{12}}{y^{12}} \times y^{24} \times 2^{12}$
$= x^{12} \times \frac{y^{24}}{y^{12}} \times 2^{12}$
$= x^{12} \times y^{24 - 12} \times 2^{12}$
$= x^{12} \times y^{12} \times 2^{12}$
$= (2 x y)^{12} [since a^{m} \times b^{m} \times c^{m} = (a \times b \times c)^{m}]$
(iii)
${(\frac{5}{2})}^{6} \times {(\frac{5}{2})}^{2}$
$= {(\frac{5}{2})}^{8} [since a^{m} \times a^{n} = a^{m + n}]$

(iv)
${(\frac{2}{3})}^{5} \times {(\frac{3}{5})}^{5}$
$= {(\frac{2}{3} \times \frac{3}{5})}^{5} [since a^{m} \times b^{m} = (a \times b)^{m}]$
$= {(\frac{2}{5})}^{5}$

Page No 6.28:

Question 4:

Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.

Answer:

We have
9 x 9 x 9 x 9 x 9 = (9)⁵ =(3²)⁵ = 3¹⁰

Page No 6.28:

Question 5:

Simplify and write each of the following in exponential form:
(i) (25)³ ÷ 5³
(ii) (81)⁵ ÷ (3²)⁵
(iii) $\frac{9^{8} \times (x^{2})^{5}}{(27)^{4} \times (x^{3})^{2}}$
(iv) $\frac{3^{2} \times 7^{8} \times 13^{6}}{21^{2} \times 91^{3}}$

Answer:

We have
(i) (25)³ ÷ 5³
= (5²)³÷ 5³
= 5⁶ ÷ 5³
= $\frac{5^{6}}{5^{3}} = 5^{6 - 3} = 5^{3}$

(ii) (81)⁵ ÷ (3²)⁵
= (3⁴)⁵ ÷ (3²)⁵
= (3)²⁰ ÷ (3)¹⁰
= $\frac{3^{20}}{3^{10}} = 3^{20 - 10} = 3^{10}$

(iii)

$\frac{9^{8} \times (x^{2})^{5}}{(27)^{4} \times (x^{3})^{2}}$
$= \frac{(3^{2})^{8} \times (x^{2})^{5}}{(3^{3})^{4} \times (x^{3})^{2}}$
$= \frac{3^{16} \times (x)^{10}}{3^{12} \times (x)^{6}}$
$= 3$ ^16-12× (x)^10-6= 3⁴× x⁴= (3x)⁴

(iv)

$\frac{3^{2} \times 7^{8} \times 13^{6}}{21^{2} \times 91^{3}}$
$= \frac{3^{2} \times 7^{2} \times 7^{6} \times 13^{6}}{21^{2} \times (13 \times 7)^{3}}$
$= \frac{(21)^{2} \times 7^{6} \times 13^{6}}{21^{2} \times 13^{3} \times 7^{3}}$
$= \frac{7^{6} \times 13^{6}}{13^{3} \times 7^{3}}$
$= \frac{91^{6}}{91^{3}} = 91^{6 - 3} = 91^{3}$

Page No 6.28:

Question 6:

Simplify:
(i) $(3^{5})^{11} \times (3^{15})^{4} - (3^{5})^{18} \times (3^{5})^{5}$
(ii) $\frac{16 \times 2^{n + 1} - 4 \times 2^{n}}{16 \times 2^{n + 2} - 2 \times 2^{n + 2}}$
(iii) $\frac{10 \times 5^{n + 1} + 25 \times 5^{n}}{3 \times 5^{n + 2} + 10 \times 5^{n + 1}}$
(iv) $\frac{(16)^{7} \times (25)^{5} \times (81)^{3}}{(15)^{7} \times (24)^{5} \times (80)^{3}}$

Answer:

We have
(i) (3⁵)¹¹× (3¹⁵)⁴- (3⁵)¹⁸×(3⁵)⁵
   = 3⁵⁵ x 3⁶⁰ - 3⁹⁰ x 3²⁵
   = 3^{(55 + 60)} - 3^{(90 + 25)}
   = 3¹¹⁵ - 3¹¹⁵
   = 0

(ii) $\frac{16 \times 2^{n + 1} - 4 \times 2^{n}}{16 \times 2^{n + 2} - 2 \times 2^{n + 2}}$
    = $\frac{2^{4} \times 2^{n + 1} - 2^{2} \times 2^{n}}{2^{4} \times 2^{n + 2} - 2^{n + 1} \times 2^{2}}$
   $= \frac{2^{2} \times (2^{n + 3} - 2^{n})}{2^{2} \times (2^{n + 4} - 2^{n + 1})}$
$= \frac{2^{n} \times 2^{3} - 2^{n}}{2^{n} \times 2^{4} - 2^{n} \times 2}$
$= \frac{2^{n} (2^{3} - 1)}{2^{n} (2^{4} - 2)} = \frac{8 - 1}{16 - 2} = \frac{7}{14} = \frac{1}{2}$

(iii)

$\frac{10 \times 5^{n + 1} + 25 \times 5^{n}}{3 \times 5^{n + 2} + 10 \times 5^{n + 1}}$
$= \frac{10 \times 5^{n + 1} + (5)^{2} \times 5^{n}}{3 \times 5^{n + 2} + 2 \times 5 \times 5^{n + 1}}$
$= \frac{10 \times 5^{n + 1} + 5 \times 5^{n + 1}}{3 \times 5^{n + 2} + 2 \times 5 \times 5^{n + 1}}$
$= \frac{5^{n + 1} (10 + 5)}{3 \times 5 \times 5^{n + 1} + 10 \times 5^{n + 1}}$
$= \frac{5^{n + 1} (15)}{5^{n + 1} (15 + 10)} = \frac{5^{n + 1} \times 15}{5^{n + 1} \times 25} = \frac{15}{25} = \frac{3}{5}$

(iv)
$\frac{(16)^{7} \times (25)^{5} \times (81)^{3}}{(15)^{7} \times (24)^{5} \times (80)^{3}}$
$= \frac{(16)^{7} \times (5^{2})^{5} \times (3^{4})^{3}}{(3 \times 5)^{7} \times (3 \times 8)^{5} \times {(16 \times 5)}^{3}}$
$= \frac{(16)^{7} \times (5)^{10} \times (3)^{12}}{3^{7} \times 5^{7} \times 3^{5} \times 8^{5} \times 16^{3} \times 5^{3}}$
$= \frac{(16)^{7} \times (5)^{10} \times (3)^{12}}{3^{7} \times 3^{5} \times 5^{7} \times 5^{3} \times 8^{5} \times 16^{3}}$
$= \frac{(16)^{7} \times (5)^{10} \times (3)^{12}}{3^{12} \times 5^{10} \times 8^{5} \times 16^{3}}$
$= \frac{(16)^{7}}{8^{5} \times 16^{3}}$
$= \frac{(16)^{7 - 3}}{8^{5}} = \frac{(16)^{4}}{8^{5}} = \frac{(2 \times 8)^{4}}{8^{5}} = \frac{2^{4} \times 8^{4}}{8^{5}} = \frac{2^{4}}{8} = \frac{16}{8} = 2$

Page No 6.28:

Question 7:

Find the values of n in each of the following:
(i) $5^{2 n} \times 5^{3} = 5^{11}$
(ii) $9 \times 3^{n} = 3^{7}$
(iii) $8 \times 2^{n + 2} = 32$
(iv) $7^{2 n + 1} \div 49 = 7^{3}$
(v) ${(\frac{3}{2})}^{4} \times {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 n + 1}$
(vi) ${(\frac{2}{3})}^{10} + {\{{(\frac{3}{2})}^{2}\}}^{5} = {(\frac{2}{5})}^{2 n - 2}$

Answer:

We have

(i) 5²ⁿ x 5³ = 5¹¹
= 5²ⁿ⁺³= 5¹¹
On equating the coefficients, we get
2n + 3 = 11
⇒2n = 11- 3
⇒2n = 8
⇒ n = $\frac{8}{2} = 4$
(ii) 9 x 3ⁿ = 3⁷
= (3)² x 3ⁿ = 3⁷
= (3)²⁺ⁿ= 3⁷
On equating the coefficients, we get
2 + n = 7
⇒ n = 7 - 2 = 5

(iii) 8 x 2ⁿ⁺² = 32
= (2)³ x 2ⁿ⁺² = (2)⁵ [since 2³ = 8 and 2⁵ = 32]
= (2)³⁺ⁿ⁺²= (2)⁵
On equating the coefficients, we get
3 + n + 2 = 5
⇒ n + 5 = 5
⇒ n = 5 -5
⇒ n = 0

(iv) 7²ⁿ⁺¹ ÷ 49 = 7³
= 7²ⁿ⁺¹ ÷ 7² = 7³ [since 49 = 7²]
$= \frac{7^{2 n + 1}}{7^{2}} = 7^{3}$
$= 7^{2 n + 1 - 2} = 7^{3} [since \frac{a^{m}}{a^{n}} = a^{m - n}]$
= 7^2n-1=7³
On equating the coefficients, we get
2n - 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
⇒ n = $\frac{4}{2} = 2$
(v) ${(\frac{3}{2})}^{4} \times {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 n + 1}$
$= {(\frac{3}{2})}^{(4 + 5)} = {(\frac{3}{2})}^{(2 n + 1)}$
$= {(\frac{3}{2})}^{9} = {(\frac{3}{2})}^{2 n + 1}$
On equating the coefficients, we get
2n + 1 = 9
⇒ 2n = 9 - 1
⇒ 2n = 8
⇒ n = $\frac{8}{2} = 4$
(vi) ${(\frac{2}{3})}^{10} \times {\{{(\frac{3}{2})}^{2}\}}^{5} = {(\frac{2}{5})}^{2 n - 2}$
$= {(\frac{2}{3})}^{10} \times {(\frac{3}{2})}^{10} = {(\frac{2}{5})}^{2 n - 2}$
$= \frac{2^{10} \times 3^{10}}{3^{10} \times 2^{10}} = {(\frac{2}{5})}^{2 n - 2}$
$= 1 = {(\frac{2}{5})}^{2 n - 2}$
$= {(\frac{2}{5})}^{0} = {(\frac{2}{5})}^{2 n - 2} [since {(\frac{2}{5})}^{0} = 1]$
On equating the coefficients, we get
⇒ 0 = 2n - 2
⇒ 2n = 2
⇒ n = $\frac{2}{2} = 1$

Page No 6.29:

Question 8:

If $\frac{9^{n} \times 3^{2} \times 3^{n} - (27)^{n}}{(3^{3}) \times 2^{3}} = \frac{1}{27}$ , find the value of n.

Answer:

We have

$\frac{9^{n} \times 3^{2} \times 3^{n} - (27)^{n}}{(3^{3})^{5} \times 2^{3}} = \frac{1}{27}$
$= \frac{(3^{2})^{n} \times 3^{2} \times 3^{n} - (3^{3})^{n}}{(3)^{15} \times 2^{3}} = \frac{1}{27}$
$= \frac{(3)^{2 n + 2 + n} - (3)^{3 n}}{(3)^{15} \times 2^{3}} = \frac{1}{27}$
$= \frac{(3)^{3 n + 2} - (3)^{3 n}}{(3)^{15} \times 2^{3}} = \frac{1}{27}$
$= \frac{(3)^{3 n} \times (3)^{2} - (3)^{3 n}}{(3)^{15} \times 2^{3}} = \frac{1}{27}$
$= \frac{(3)^{3 n} (3^{2} - 1)}{(3)^{15} \times 2^{3}} = \frac{1}{27}$
$= \frac{(3)^{3 n} \times 8}{(3)^{15} \times 2^{3}} = \frac{1}{27}$
$= \frac{(3)^{3 n} \times 2^{3}}{(3)^{15} \times 2^{3}} = \frac{1}{27}$
$= \frac{3^{3 n}}{3^{15}} = \frac{1}{27}$
$= 3^{3 n - 15} = \frac{1}{3^{3}}$
$= 3^{3 n - 15} = 3^{- 3}$
On equating the coefficients, we get
3n -15 = -3
⇒ 3n = -3 + 15
⇒ 3n = 12
⇒ n = $\frac{12}{3} = 4$

Page No 6.30:

Question 1:

Express the following numbers in the standard form:
(i) 3908.78
(ii) 5,00,00,000
(iii) 3,18,65,00,000
(iv) 846 × 10⁷
(v)723 × 10⁹

Answer:

We have

(i) 3908.78 = 3.90878 x 10³                 [since the decimal point is moved 3 places to the left]

(ii) 5,00,00,000 = 5,00,00,000.00 = 5 x 10⁷               [since the decimal point is moved 7 places to the left]

(iii) 3,18,65,00,000 = 3,18,65,00,000.00
                                = 3.1865 x 10⁹                            [since the decimal point is moved 9 places to the left]

(iv) 846 × 10⁷ = 8.46 x 10² x 10⁷         [since the decimal point is moved 2 places to the left]
                        = 8.46 x 10⁹                                        [since a^m x aⁿ = a^m+n]

(v) 723 × 10⁹ = 7.23 x 10² x 10⁹                               [since the decimal point is moved 2 places to the left]
                      = 7.23 x 10¹¹                                        [ since a^m x aⁿ = a^m+n]

Page No 6.30:

Question 2:

Write the following numbers in the usual form:
(i) 4.83 × 10⁷
(ii) 3.21 × 10⁵
(iii) 3.5 × 10³

Answer:

We have
(i) 4.83 × 10⁷ = 483 × 10^7-2                        [since the decimal point is moved two places to the right]
                      = 483 × 10⁵ = 4,83,00,000

(ii) 3.21 × 10⁵= 321 x 10^5-2                       [since the decimal point is moved two places to the right]
                      = 321 x 10³ = 3,21,000

(iii) 3.5 × 10³ = 35 x 10^3-1                          [since the decimal point is moved one place to the right]
                       = 35 x 10² = 3,500

Page No 6.30:

Question 3:

Express the numbers appearing in the following statements in the standard form:
(i) The distance between the Earth and the Moon is 384,000,000 metres.
(ii) Diameter of the Earth is 1,27,56,000 metres.
(iii) Diameter of the Sun is 1,400,000,000 metres.
(iv) The universe is estimated to be about 12,000,000,000 years old.

Answer:

We have
(i) The distance between the Earth and the Moon is 3.84 x 10⁸ metres.
     [Since the decimal point is moved 8 places to the left.]

(ii) The diameter of the Earth is 1.2756 x 10⁷ metres.
      [Since the decimal point is moved 7 places to the left.]

(iii) The diameter of the Sun is 1.4 x 10⁹ metres.
       [Since the decimal point is moved 9 places to the left.]

(iv) The universe is estimated to be about 1.2x 10¹⁰ years old.
       [Since the decimal point is moved 10 places to the left.]

Page No 6.31:

Question 1:

Write the following numbers in the expanded exponential forms:
(i) 20068
(ii) 420719
(iii) 7805192
(iv) 5004132
(v) 927303

Answer:

We have
(i) 20068 = 2 x 10⁴ + 0 x 10³ + 0 x 10² + 6 x 10¹ + 8 x 10⁰
(ii) 420719 = 4 x 10⁵ + 2 x 10⁴ + 0 x 10³ + 7 x 10² + 1 x 10¹ + 9 x 10⁰
(iii) 7805192 = 7 x 10⁶ + 8 x 10⁵ + 0 x 10⁴ + 5 x 10³ + 1 x 10² + 9 x 10¹ + 2 x 10⁰
(iv) 5004132 = 5 x 10⁶ + 0 x 10⁵ + 0 x 10⁴ 4 x 10³ + 1 x 10² + 3 x 10¹ + 2 x 10⁰
(v) 927303 = 9 x 10⁵ + 2 x 10⁴ + 7 x 10³ + 3 x 10² + 0 x 10¹ + 3 x 10⁰

Note: a⁰ = 1

Page No 6.31:

Question 2:

Find the number from each of the following expanded forms:
(i) 7 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
(ii) 5 × 10⁵ + 4 × 10⁴ + 2 × 10³ + 3 × 10⁰
(iii) 9 × 10⁵ + 5 × 10² + 3 × 10¹
(iv) 3 × 10⁴+ 4 × 10² + 5 × 10⁰

Answer:

We have

(i) 7 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
    = 7 x 10000 + 6 x 1000 + 0 x 100 + 4 x 10 + 5 x 1 = 76045

(ii) 5 × 10⁵ + 4 × 10⁴ + 2 × 10³ + 3 × 10⁰
      = 5 x 100000 + 4 x 10000 + 2 x 1000 + 3 x 1 = 542003

(iii) 9 × 10⁵ + 5 × 10² + 3 × 10¹= 9 x 100000 + 5 x 100 + 3 x 10 = 900530

(iv) 3 × 10⁴+ 4 × 10² + 5 × 10⁰
     = 3 x 10000 + 4 x 100 + 5 x 1 = 30405

Page No 6.32:

Question 1:

Mark the correct alternative in the following question:

${(6^{- 1} - 8^{- 1})}^{- 1} = (a) \frac{1}{24} (b) 24 (c) - 24 (d) - \frac{1}{24}$

Answer:

${(6^{- 1} - 8^{- 1})}^{- 1} = {(\frac{1}{6} - \frac{1}{8})}^{- 1} (As, x^{- 1} = \frac{1}{x}) = {(\frac{4 - 3}{24})}^{- 1} = {(\frac{1}{24})}^{- 1} = \frac{24}{1} (As, x^{- 1} = \frac{1}{x}) = 24$

Hence, the correct alternative is option (b).

Page No 6.32:

Question 2:

Mark the correct alternative in the following question:

$2^{3^{2}} = (a) 64 (b) 32 (c) 256 (d) 512$

Answer:

Since, $2^{3^{2}} = 2^{9} = 512$

Hence, the correct alternative is option (d).

Page No 6.32:

Question 3:

Mark the correct alternative in the following question:

${(3^{- 1} \times 5^{- 1})}^{- 1} = (a) \frac{1}{15} (b) - \frac{1}{15} (c) 15 (d) - 15$

Answer:

${(3^{- 1} \times 5^{- 1})}^{- 1} = {(\frac{1}{3} \times \frac{1}{5})}^{- 1} (As, x^{- 1} = \frac{1}{x}) = {(\frac{1}{15})}^{- 1} = \frac{15}{1} (As, x^{- 1} = \frac{1}{x}) = 15$

Hence, the correct alternative is option (c).

Page No 6.32:

Question 4:

Mark the correct alternative in the folowing question:

${(- \frac{3}{5})}^{- 1} = (a) \frac{3}{5} (b) \frac{5}{3} (c) - \frac{5}{3} (d) - \frac{3}{5}$

Answer:

Since, ${(- \frac{3}{5})}^{- 1} = - \frac{5}{3} (As, x^{- 1} = \frac{1}{x})$

Hence, the correct alternative is option (c).

Page No 6.32:

Question 5:

Mark the correct alternative in the folowing question:

${(- 1)}^{301} + {(- 1)}^{302} + {(- 1)}^{303} + . . . + {(- 1)}^{400} (a) 1 (b) 101 (c) 100 (d) 0$

Answer:

Since,

${(- 1)}^{301} + {(- 1)}^{302} + {(- 1)}^{303} + . . . + {(- 1)}^{400} = (- 1) + (1) + (- 1) + . . . + (1) [As, {(- 1)}^{odd} = - 1 and {(- 1)}^{even} = 1] = - 50 + 50 [As, there are 50 (- 1)' s and 50 (1)' s] = 0$

Hence, the correct alternative is option is (d).

Page No 6.32:

Question 6:

Mark the correct alternative in the folowing question:

$If a = 25, then a^{25^{0}} + a^{0^{25}} = (a) 25 (b) 26 (c) 24 (d) 0$

Answer:

Since,

$a^{25^{0}} + a^{0^{25}} = a^{1} + a^{0} (As, 25^{0} = 1 and 0^{25} = 0) = a + 1 (As, a^{0} = 1) = 25 + 1 = 26$

Hence, the correct alternative is option (b).

Page No 6.32:

Question 7:

Mark the correct alternative in the following question:

$\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3} = (a) \frac{19}{64} (b) \frac{64}{19} (c) \frac{27}{16} (d) \frac{- 19}{64}$

Answer:

Since,

$\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3} = \{{(\frac{3}{1})}^{3} - {(\frac{2}{1})}^{3}\} \div {(\frac{4}{1})}^{3} (As, x^{- 1} = \frac{1}{x}) = \{3^{3} - 2^{3}\} \div 4^{3} = \{27 - 8\} \div 64 = 19 \div 64 = \frac{19}{64}$

Hence, the correct alternative is option (a).

Page No 6.32:

Question 8:

Mark the correct alternative in the following question:

${(25^{2} - 15^{2})}^{\frac{3}{2}} = (a) 4000 (b) 8000 (c) 3125 (d) 1024$

Answer:

Since,

${(25^{2} - 15^{2})}^{\frac{3}{2}} = {(625 - 225)}^{\frac{3}{2}} = {(400)}^{\frac{3}{2}} = {(20^{2})}^{\frac{3}{2}} = 20^{2 \times \frac{3}{2}} [As, {(x^{m})}^{n} = x^{m n}] = 20^{3} = 8000$

Hence, the correct alternative is option (b).

Page No 6.32:

Question 9:

Mark the correct alternative in the following question:

$If {(\frac{5}{3})}^{- 5} \times {(\frac{5}{3})}^{11} = {(\frac{5}{3})}^{8 x}, then x = ? (a) - \frac{1}{2} (b) - \frac{3}{4} (c) \frac{3}{4} (d) \frac{4}{3}$

Answer:

Since,

${(\frac{5}{3})}^{- 5} \times {(\frac{5}{3})}^{11} = {(\frac{5}{3})}^{8 x} \Rightarrow {(\frac{5}{3})}^{- 5 + 11} = {(\frac{5}{3})}^{8 x} (As, x^{m} \times^{n} = x^{m + n}) \Rightarrow {(\frac{5}{3})}^{6} = {(\frac{5}{3})}^{8 x} Comparing the exponent of both the sides, we get 8 x = 6 \Rightarrow x = \frac{6}{8} ∴ x = \frac{3}{4}$

Hence, the correct alternative is option (c).

Page No 6.32:

Question 10:

Mark the correct alternative in the following question:

${[{\{{(- \frac{1}{3})}^{2}\}}^{- 2}]}^{- 1} = (a) \frac{1}{81} (b) \frac{1}{9} (c) - \frac{1}{81} (d) - \frac{1}{9}$

Answer:

${[{\{{(- \frac{1}{3})}^{2}\}}^{- 2}]}^{- 1} = [{\{{(\frac{- 1}{3})}^{2}\}}^{(- 2) \times (- 1)}] [As, {(x^{m})}^{n} = x^{m n}] = {(\frac{- 1}{3})}^{2 \times 2} [As, {(x^{m})}^{n} = x^{m n}] = {(\frac{- 1}{3})}^{4} = \frac{{(- 1)}^{4}}{3^{4}} [As, {(\frac{x}{y})}^{m} = \frac{x^{m}}{y^{m}}] = \frac{1}{81}$

Hence, the correct alternative is option (a).

Page No 6.32:

Question 11:

Mark the correct alternative in the following question:

$\frac{{(144)}^{\frac{1}{2}} + {(256)}^{\frac{1}{2}}}{3^{2} - 2} = (a) 8 (b) 4 (c) - 4 (d) - 8$

Answer:

Since,

$\frac{{(144)}^{\frac{1}{2}} + {(256)}^{\frac{1}{2}}}{3^{2} - 2} = \frac{{(12^{2})}^{\frac{1}{2}} + {(16^{2})}^{\frac{1}{2}}}{9 - 2} = \frac{12^{2 \times \frac{1}{2}} + 16^{2 \times \frac{1}{2}}}{7} [As, {(x^{m})}^{n} = x^{m n}] = \frac{12 + 16}{7} = \frac{28}{7} = 4$

Hence, the correct alternative is option (b).

Page No 6.33:

Question 12:

Mark the correct alternative in the following question:

${(1 + 3 + 5 + 7 + 9 + 11)}^{\frac{3}{2}} = (a) 36 (b) 216 (c) 256 (d) None of these$

Answer:

Since,

${(1 + 3 + 5 + 7 + 9 + 11)}^{\frac{3}{2}} = {(36)}^{\frac{3}{2}} = {(6^{2})}^{\frac{3}{2}} = 6^{2 \times \frac{3}{2}} [As, {(x^{m})}^{n} = x^{m n}] = 6^{3} = 216$

Hence, the correct alternative is option (b).

Page No 6.33:

Question 13:

Mark the correct alternative in the following question:

$If a b c = 0, then \frac{{\{{(x^{a})}^{b}\}}^{c}}{{\{{(x^{b})}^{c}\}}^{a}} = (a) 3 (b) 0 (c) - 1 (d) 1$

Answer:

$\frac{{\{{(x^{a})}^{b}\}}^{c}}{{\{{(x^{b})}^{c}\}}^{a}} = \frac{{(x^{a})}^{b c}}{{(x^{b})}^{a c}} [As, {(x^{m})}^{n} = x^{m n}] = \frac{x^{a b c}}{x^{a b c}} [As, {(x^{m})}^{n} = x^{m n}] = \frac{x^{0}}{x^{0}} [As, a b c = 0] = \frac{1}{1} [As, x^{0} = 1] = 1$

Page No 6.33:

Question 14:

Mark the correct alternative in the following question:

${(2^{3})}^{4} = (a) 2^{4^{3}} (b) 2^{3^{4}} (c) {(2^{4})}^{3} (d) None of these$

Answer:

Since,

${(2^{3})}^{4} = 2^{3 \times 4} [As, {(x^{m})}^{n} = x^{m n}] = 2^{4 \times 3} = {(2^{4})}^{3} [As, x^{m n} = {(x^{m})}^{n}]$

Hence, the correct alternative is option (c).

Page No 6.33:

Question 15:

Mark the correct alternative in the following question:

${\{{(33)}^{2} - {(31)}^{2}\}}^{\frac{5}{7}} = (a) 64 (b) 16 (c) 32 (d) 4$

Answer:

Since,

${\{{(33)}^{2} - {(31)}^{2}\}}^{\frac{5}{7}} = {\{1089 - 961\}}^{\frac{5}{7}} = {\{128\}}^{\frac{5}{7}} = {\{2^{7}\}}^{\frac{5}{7}} = 2^{7 \times \frac{5}{7}} [As, {(x^{m})}^{n} = x^{m n}] = 2^{5} = 32$

Hence, the correct alternative is option (c).

Page No 6.33:

Question 16:

Mark the correct alternative in the following question:

$If a b c = 0, then find the value of {\{{(x^{a})}^{b}\}}^{c} . (a) 1 (b) a (c) b (d) c$

Answer:

Since,

${\{{(x^{a})}^{b}\}}^{c} = {(x^{a})}^{b c} [As, {(x^{m})}^{n} = x^{m n}] = x^{a b c} [As, {(x^{m})}^{n} = x^{m n}] = x^{0} [As, a b c = 0] = 1 [As, x^{0} = 1]$

Hence, the correct alternative is option (a).

Page No 6.33:

Question 17:

Mark the correct alternative in the following question:

$If a = 3^{- 3} - 3^{3} and b = 3^{3} - 3^{- 3}, then \frac{a}{b} - \frac{b}{a} = (a) 0 (b) 1 (c) - 1 (d) 2$

Answer:

$Since, a = 3^{- 3} - 3^{3} = \frac{1}{3^{3}} - 3^{3} (As, x^{- m} = \frac{1}{x^{m}}) = \frac{1}{27} - \frac{27}{1} = \frac{1}{27} - \frac{27 \times 27}{1 \times 27} = \frac{1}{27} - \frac{729}{27} = \frac{1 - 729}{27} = \frac{- 728}{27} Also, b = 3^{3} - 3^{- 3} = 3^{3} - \frac{1}{3^{3}} (As, x^{- m} = \frac{1}{x^{m}}) = \frac{27}{1} - \frac{1}{27} = \frac{27 \times 27}{1 \times 27} - \frac{1}{27} = \frac{729}{27} - \frac{1}{27} = \frac{729 - 1}{27} = \frac{728}{27} Now, \frac{a}{b} - \frac{b}{a} = \frac{(\frac{- 728}{27})}{(\frac{728}{27})} - \frac{(\frac{728}{27})}{(\frac{- 728}{27})} = (- 1) - (- 1) = - 1 + 1 = 0$

Hence, the correct alternative is option (a).

Page No 6.33:

Question 18:

Mark the correct alternative in the following question:

$What should be multiplied to 6^{- 2} so that the product may be equal to 216 ? (a) 6^{4} (b) 6^{5} (c) 6^{3} (d) 6$

Answer:

$Let 6^{m} should be multiplied to 6^{- 2} . A . T . Q . 6^{- 2} \times 6^{m} = 216 \Rightarrow 6^{- 2 + m} = 6^{3} Comparing th exponent of both the sides, we get - 2 + m = 3 \Rightarrow m = 3 + 2 ∴ m = 5$

So, 6⁵ should be multiplied.

Hence, the correct alternative is option (b).

Page No 6.33:

Question 19:

Mark the correct alternative in the following question:

$If x y z = 0, then find the value of {(a^{x})}^{y z} + {(a^{y})}^{z x} + {(a^{z})}^{x y} = (a) 3 (b) 2 (c) 1 (d) 0$

Answer:

Since,

${(a^{x})}^{y z} + {(a^{y})}^{z x} + {(a^{z})}^{x y} = a^{x y z} + a^{y z x} + a^{z x y} [As, {(x^{m})}^{n} = x^{m n}] = a^{x y z} + a^{x y z} + a^{x y z} = a^{0} + a^{0} + a^{0} = 1 + 1 + 1 (As, x^{0} = 1) = 3$

Hence, the correct option is (a).

Page No 6.33:

Question 20:

Choose the correct alternative in the following question:

$If 2^{n} = 4096, then 2^{n - 5} = (a) 128 (b) 64 (c) 256 (d) 32$

Answer:

$As, 2^{n} = 4096 \Rightarrow 2^{n} = 2^{12} Comparing the exponent of both the sides, we get n = 12 Now, 2^{n - 5} = 2^{12 - 5} = 2^{7} = 128$

Hence, the correct alternative is option (a).

Page No 6.33:

Question 21:

Choose the correct alternative in the following question:

The number 4,70,394 in standard form is written as

(a) 4.70394 $\times$ 10⁵ (b) 4.70394 $\times$ 10⁴ (c) 47.0394 $\times$ 10⁴ (d) 4703.94 $\times$ 10²

Answer:

Since, 4,70,394 = 4.70394 $\times$ 100000 = 4.70394 $\times$ 10⁵.

So, the number 4,70,394 in standard form is written as 4.70394 $\times$ 10⁵.

Hence, the correct alternative is option (a).

Page No 6.33:

Question 22:

Choose the correct alternative in the following question:

The number 2.35 $\times$ 10⁴ in the usual form is written as

(a) 2.35 $\times$ 10³ (b) 23500 (c) 2350000 (d) 235 $\times$ 10⁴

Answer:

Since, 2.35 $\times$ 10⁴ = 2.35 $\times$ 10000 = 23500

So, the number 2.35 $\times$ 10⁴ in the usual form is written as 23500.

Hence, the correct alternative is option (b).

Page No 6.33:

Question 23:

Choose the correct alternative in the following question:

$If 3^{x} = 6561, then 3^{x - 3} = (a) 81 (b) 243 (c) 729 (d) 27$

Answer:

$As, 3^{x} = 6561 \Rightarrow 3^{x} = 3^{8} Comparing the exponent of both the sides, we get x = 8 Now, 3^{x - 3} = 3^{8 - 3} = 3^{5} = 243$

Hence, the correct alternative is option (b).

Page No 6.33:

Question 24:

Choose the correct alternative in the following question:

$If 2^{n} = 1024, then 2^{(\frac{n}{2} + 2)} = (a) 64 (b) 128 (c) 256 (d) 512$

Answer:

$As, 2^{n} = 1024 \Rightarrow 2^{n} = 2^{10} Comparing the exponent of both the sides, we get n = 10 Now, 2^{(\frac{n}{2} + 2)} = 2^{(\frac{10}{2} + 2)} = 2^{(5 + 2)} = 2^{7} = 128$

Hence, the correct alternative is option (b).

Page No 6.33:

Question 25:

Choose the correct alternative in the following question:

${(8^{4} + 8^{2})}^{\frac{1}{2}} = (a) 84 (b) 8 \sqrt{77} (c) 72 (d) 8 \sqrt{65}$

Answer:

Since,

${(8^{4} + 8^{2})}^{\frac{1}{2}} = {(8^{2 + 2} + 8^{2})}^{\frac{1}{2}} = {(8^{2} \times 8^{2} + 8^{2})}^{\frac{1}{2}} (As, x^{m + n} = x^{m} \times x^{n}) = {[8^{2} \times (8^{2} + 1)]}^{\frac{1}{2}} [As, a b + a c = a \times (b + c)] = {(8^{2})}^{\frac{1}{2}} \times {(8^{2} + 1)}^{\frac{1}{2}} [As, {(a b)}^{m} = a^{m} \times b^{m}] = 8^{2 \times \frac{1}{2}} \times {(64 + 1)}^{\frac{1}{2}} = 8 \times 65^{\frac{1}{2}} = 8 \sqrt{65}$

Hence, the correct alternative is option (d).

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