RD Sharma solution class 7 chapter 6 Exponents Exercise 6.1

Exercise 6.1

Page-6.12

Question 1:

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

Answer 1:

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

Question 2:

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

Answer 2:

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

Question 3:

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

Answer 3:

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]

Question 4:

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

Answer 4:

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]

Question 5:

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

Answer 5:

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]

Question 6:

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

Answer 6:

We have

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

Question 7:

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

Answer 7:

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.

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 8:

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

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

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

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) $\left(\frac{-2}{3}\right)\times \left(\frac{-2}{3}\right)\times x\times x\times x$

Answer 9:

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) $\left(\frac{-2}{3}\right)\times \left(\frac{-2}{3}\right)\times x\times x\times x = {\left(\frac{-2}{3}\right)}^2\times x^3$

Question 10:

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

Answer 10:

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

Question 11:

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

Answer 11:

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-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 12:

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

Question 13:

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

Answer 13:

We have

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

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

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 14:

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

Question 15:

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

Answer 15:

We have

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

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

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) ${\left(\frac{b}{a}\right)}^b$
(iv) ${\left(\frac{a}{b}+\frac{b}{a}\right)}^a$

Answer 16:

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

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