Number Systems ⇒⇒ Exercise 1.5

Question 1 (i)

Find:
(i) $ {64^{1 \over 2} } $

Solution :

Given : $ {64^{1 \over 2} } $

Now, on simplifying it, we will get prime factors of 64 = 2 × 2 × 2 × 2 × 2 × 2

Hence, $$({ 2 × 2 × 2 × 2 × 2 × 2 })^{1 \over 2} $$

$$⇒ ({ 2^6})^{1 \over 2} $$

(According to Laws of Exponents = ${(a^m )^n } = a^{m × n}$)

$$⇒ (2)^{ 6 ×{ 1 \over 2}} $$

$$⇒ 2^3 $$

$$⇒ 8 $$

Question 1 (ii)

Find:
(ii) $ {32^{1 \over 5} } $

Solution :

Given : $ {32^{1 \over 5} } $

Now, on simplifying it, we will get prime factors of 32 = 2 × 2 × 2 × 2 × 2

Hence, $$({ 2 × 2 × 2 × 2 × 2 })^{1 \over 5} $$

$$⇒ ({ 2^5})^{1 \over 5} $$

(According to Laws of Exponents = ${(a^m )^n } = a^{m × n}$ )

$$⇒ (2)^{ 5 ×{ 1 \over 5}} $$

$$⇒ 2^1 $$

$$⇒ 2 $$

Question 1 (iii)

Find:
(iii) $ {125^{1 \over 3} } $

Solution :

Given : $ {125^{1 \over 3} } $

Now, on simplifying it, we will get prime factors of 125 = 5 × 5 × 5

Hence, $$({ 5 × 5 × 5 })^{1 \over 3} $$

$$⇒ ({ 5^3})^{1 \over 3} $$

(According to Laws of Exponents = ${(a^m )^n } = a^{m × n}$)

$$⇒ (5)^{ 3 ×{ 1 \over 3}} $$

$$⇒ 5^1 $$

$$⇒ 5 $$

Question 2 (i)

Find:
(i) $ {9^{3 \over 2} } $

Solution :

Given : $ {9^{3 \over 2} } $

Now, on simplifying it, we will get prime factors of 9 = 3 × 3

Hence, $$({ 3 × 3})^{3 \over 2} $$

$$⇒ ({ 3^2})^{3 \over 2} $$

(According to Laws of Exponents = ${(a^m )^n } = a^{m × n}$)

$$⇒ (3)^{ 2 ×{ 3 \over 2}} $$

$$⇒ 3^3 $$

$$⇒ 3 × 3 × 3 = 27 $$

Question 2 (ii)

Find:
(ii) $ {32^{2 \over 5} } $

Solution :

Given : $ {32^{2 \over 5} } $

Now, on simplifying it, we will get prime factors of 32 = 2 × 2 × 2 × 2 × 2

Hence, $$({ 2 × 2 × 2 × 2 × 2 })^{2 \over 5} $$

$$⇒ ({ 2^5})^{2 \over 5} $$

(According to Laws of Exponents = ${(a^m )^n } = a^{m × n}$)

$$⇒ (2)^{ 5 ×{ 2 \over 5}} $$

$$⇒ 2^2 $$

$$⇒ 2 × 2 $$

$$⇒ 4 $$

Question 2 (iii)

Find:
(iii) $ {16^{3 \over 4} } $

Solution :

Given : $ {16^{3 \over 4} } $

Now, on simplifying it, we will get prime factors of 16 = 2 × 2 × 2 × 2

Hence, $$({ 2 × 2 × 2 × 2 })^{3 \over 4} $$

$$⇒ ({ 2^4})^{3 \over 4} $$

(According to Laws of Exponents = ${(a^m )^n } = a^{m × n}$)

$$⇒ (2)^{ 4 ×{ 3 \over 4}} $$

$$⇒ 2^3 $$

$$⇒ 2 × 2 × 2 $$

$$⇒ 8 $$

Question 2 (iv)

Find:
(iv) $ {125^{-1 \over 3} } $

Solution :

Given : $ {125^{-1 \over 3} } $

Now, on simplifying it, we will get prime factors of 125 = 5 × 5 × 5

Hence, $$({ 5 × 5 × 5 })^{-1 \over 3} $$

$$⇒ ({ 5^3})^{-1 \over 3} $$

(According to Laws of Exponents = ${(a^m )^n } = a^{m × n}$)

$$⇒ (5)^{ 3 ×{ -1 \over 3}} $$

$$⇒ 5^{-1} $$

( Now, According to Laws of Exponents = ${a^{-m} } = { 1 \over {a^m}}$)

$$⇒ { 1 \over 5}$$

Question 3 (i)

Find:
(i) $ {2^{2 \over 3} } . {2^{1 \over 5} }$

Solution :

Given : $ {2^{2 \over 3} } . {2^{1 \over 5} }$

Hence, $$( {2^{2 \over 3} } × {2^{1 \over 5} }) $$

(According to Laws of Exponents = $ {a^m × a^n } = a^{m + n}$)

$$⇒ (2)^{{2 \over 3} +{1 \over 5}} $$

$$⇒ (2)^{{10 + 3} \over 15} $$

$$⇒ (2)^{13 \over 15} $$

Question 3 (ii)

Find:
(ii) $ { ({1 \over {3^3}})^7 }$

Solution :

Given : $ { ({1 \over {3^3}})^7 }$

( According to Laws of Exponents = ${a^{-m} } = { 1 \over {a^m}}$ )

$$⇒ { ({3^{-3}})^7 } $$

( Now, According to Laws of Exponents = ${(a^m )^n } = a^{m × n}$)

$$⇒ (3)^{{-3} × 7} $$

$$⇒ (3)^{-21} $$

Question 3 (iii)

Find:
(iii) $ {{11^ {1 \over 2}} \over {{11^ {1 \over 4}}}} $

Solution :

Given : $ {{11^ {1 \over 2}} \over {{11^ {1 \over 4}}}} $

( According to Laws of Exponents = $ {a^m \over { a^n} } = a^{m - n}$)

$$⇒ {(11)^{{1 \over 2}- {1 \over 4}}} $$

$$⇒ {(11)^{{2 - 1} \over 4}} $$

$$⇒ {11^ {1 \over 4}} $$

Question 3 (iv)

Find:
(iv) $ {7^{1 \over 2} } . {8^{1 \over 2} }$

Solution :

Given : $ {7^{1 \over 2} } . {8^{1 \over 2} }$

( According to Laws of Exponents = $ {a^m × b^m } = {(ab)^m} $)

$$⇒ {(7 × 8)^{1 \over 2} } $$

$$⇒ {(56)^{1 \over 2} } $$

$$⇒ {(56)^{1 \over 2} } $$