Exponents and Powers

EXERCISE-13.1

P1.Find the value of:

(i) 2⁶

(ii) 9³

(iii) 11²

(iv)5⁴

Sol. (i) 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9³ = 9 × 9 × 9 = 729

(iii) 11² = 11 × 11 = 121

(iv)5⁴ = 5 × 5 × 5 × 5 = 625

P2. Express the following in exponential form:

(i) 6 × 6 × 6 × 6

(ii) t × t

(iii) b × b × b × b

(iv) 5 × 5 × 7 ×7 × 7

(v) 2 × 2 × a × a

(vi) a × a × a × c × c × c × c × d

Sol. (i) 6 × 6 × 6 × 6 = 6⁴

(ii) t × t= t²

(iii) b × b × b × b = b⁴

(iv) 5 × 5 × 7 × 7 × 7 = 5² × 7³

(v) 2 × 2 × a × a = 2² × a²

(vi) a × a × a × c × c × c × c × d  = a³ c⁴ d

P3.Express the following numbers using exponential notation:

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Sol. (i) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹

(ii) 343 = 7 × 7 × 7 = 7³

(iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶

(iv) 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

P4. Identify the greater number, wherever possible, in each of the following?

(i) 4³ or 3⁴

(ii) 5³ or 3⁵

(iii) 2⁸ or 8²

(iv) 100² or 2¹⁰⁰

(v) 2¹⁰ or 10²

Sol. (i) 4³ = 4 × 4 × 4 = 64

3⁴ = 3 × 3 × 3 × 3 = 81

Therefore, 3⁴ > 4³

(ii)  5³ = 5 × 5 × 5 =125

3⁵ = 3 × 3 × 3 × 3 × 3 = 243

Therefore, 3⁵ > 5³

(iii) 2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

8² = 8 × 8 = 64

Therefore, 2⁸ > 8²

(iv) 100² or 2¹⁰⁰

2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

2¹⁰⁰ = 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 ×1024 × 1024

100² = 100 × 100 = 10000

Therefore, 2¹⁰⁰ > 100²

(v) 2¹⁰ and 10²

2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

10² = 10 × 10 = 100

Therefore, 2¹⁰ > 10²

P5. Express each of the following as product of powers of their prime factors:

(i) 648

(ii) 405

(iii) 540

(iv) 3,600

Sol. (i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2³. 3⁴

(ii) 405 = 3 × 3 × 3 × 3 × 5 = 3⁴ . 5

(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 2². 3³. 5

(iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴. 3². 5²

P6. Simplify:

(i) 2 × 10³       (ii) 7² × 2²

(iii) 2³ × 5      (iv) 3 × 4⁴

(v) 0 × 10² _­­­­­     (vi) 5² × 3³

(vii) 2⁴ × 3²    (viii) 3² × 10⁴

Sol. (i) 2 × 10³ = 2 × 10 × 10 × 10 = 2 × 1000 = 2000

(ii) 7² × 2² = 7 × 7 × 2 × 2 = 49 × 4 = 196

(iii) 2³ × 5 = 2 × 2 × 2 × 5 = 8 × 5 = 40

(iv) 3 × 4⁴ = 3 × 4 × 4 × 4 × 4 = 3 × 256 = 768

(v) 0 × 10² = 0 × 10 × 10 = 0

(vi) 5² × 3³ = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675

(vii) 2⁴ × 3² = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144

(viii) 3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

P7. Simplify:

(i) (− 4)³

(ii) (− 3) × (− 2)³

(iii) (− 3)² × (− 5)²

(iv)(− 2)³ × (−10)³

Sol. (i) (−4)³ = (−4) × (−4) × (−4) = −64

(ii) (−3) × (−2)³ = (−3) × (−2) × (−2) × (−2) = 24

(iii) (−3)² × (−5)² = (−3) × (−3) × (−5) × (−5) = 9 × 25 = 225

(iv) (−2)³ × (−10)³ = (−2) × (−2) × (−2) × (−10) × (−10) × (−10) = (−8) × (−1000) = 8000

P8.Compare the following numbers:

(i) 2.7 × 10¹²; 1.5 × 10⁸

(ii) 4 × 10¹⁴; 3 × 10¹⁷

Sol. (i) 2.7 × 10¹²; 1.5 × 10⁸

2.7 × 10¹² > 1.5 × 10⁸

(ii) 4 × 10¹⁴; 3 × 10¹⁷

3 × 10¹⁷ > 4 × 10¹⁴

EXERCISE-12.2

P1. Using laws of exponents, simplify and write the answer in exponential form:

(i) 3² × 3⁴ × 3⁸           (ii) 6¹⁵ ÷ 6¹⁰

(iii) a³ × a²                       (iv) 7^x× 7²

(v)  (vi) 2⁵ × 5⁵              (vii) a⁴ × b⁴

(viii) (3⁴)³                          (ix) $\left(2^{20}÷2^{15}\right)\times 2^3$

(x) 8^t ÷ 8²

Sol. (i)3² × 3⁴ × 3⁸ = (3)^{2 + 4 + 8} (a^m × aⁿ = a^m+n) = 3¹⁴

(ii) 6¹⁵ ÷ 6¹⁰ = (6)^{15 − 10} (a^m ÷ aⁿ = a^m−n) = 6⁵

(iii) a³ × a² = a^{(3 + 2)} (a^m × aⁿ = a^m+n) = a⁵

(iv) 7^x + 7² = 7^{x + 2} (a^m × aⁿ = a^m+n)

(v) (5²)³ ÷ 5³

= 5^{2 × 3} ÷ 5³ (a^m)ⁿ = a^mn

= 5⁶ ÷ 5³

= 5^{(6 − 3)} (a^m ÷ aⁿ = a^m−n)

= 5³

(vi) 2⁵ × 5⁵ = (2 × 5)⁵ [a^m × b^m = (a × b)^m] = 10⁵

(vii) a⁴ × b⁴ = (ab)⁴ [a^m × b^m = (a × b)^m]

(viii) (3⁴)³ = 3^{4 × 3} = 3¹² (a^m)ⁿ = a^mn

(ix) (2²⁰ ÷ 2¹⁵) × 2³

= (2^{20 − 15}) × 2³ (a^m ÷ aⁿ = a^m−n)

= 2⁵ × 2³

= (2^{5 + 3}) (a^m × aⁿ = a^m+n)

= 2⁸

(x) 8^t ÷ 8² = 8^{(t − 2)} (a^m ÷ aⁿ = a^m−n)

P2. Simplify and express each of the following in exponential form:

(i)  $\frac{{\mathbf{2}}^{\mathbf{3}}\mathbf{\times }{\mathbf{3}}^{\mathbf{4}}\mathbf{\times }\mathbf{4}}{\mathbf{3}\mathbf{\times }\mathbf{32}}$              (ii) $[5^2\times 5^4]\mathbf{÷}{\mathbf{5}}^{\mathbf{7}}$

(iii) ${\mathbf{25}}^{\mathbf{4}}\mathbf{÷}{\mathbf{5}}^{\mathbf{3}}$                  (iv) $\frac{\mathbf{3}\mathbf{\times }{\mathbf{7}}^{\mathbf{2}}\mathbf{\times }{\mathbf{11}}^{\mathbf{8}}}{\mathbf{21}\mathbf{\times }{\mathbf{11}}^{\mathbf{3}}}$

(v) $\frac{{\mathbf{3}}^{\mathbf{7}}}{{\mathbf{3}}^{\mathbf{4}}\mathbf{\times }{\mathbf{3}}^{\mathbf{3}}}$                     (vi) 2⁰ + 3⁰ + 4⁰

(vii) 2⁰ × 3⁰ × 4⁰             (viii) (3⁰ + 2⁰) × 5⁰

(ix) $\frac{{\mathbf{2}}^{\mathbf{8}}\mathbf{\times }{\mathbf{a}}^{\mathbf{5}}}{{\mathbf{4}}^{\mathbf{3}}\mathbf{\times }{\mathbf{a}}^{\mathbf{3}}}$                    (x) $\left(\frac{a^5}{a^3}\right)\mathbf{\times }{\mathbf{a}}^{\mathbf{8}}$

(xi) $\frac{{\mathbf{4}}^{\mathbf{5}}\mathbf{\times }{\mathbf{a}}^{\mathbf{8}}{\mathbf{b}}^{\mathbf{3}}}{{\mathbf{4}}^{\mathbf{3}}\mathbf{\times }{\mathbf{a}}^{\mathbf{3}}}$                (xii) ${(2^3\times 2)}^{\mathbf{2}}$

Sol. (i) $\frac{2^3\times 3^4\times 4}{3\times 32}=\frac{2^3\times 3^4\times 2\times 2}{3\times 2\times 2\times 2\times \times 2\times 2}$

$=\frac{2^3\times 3^4\times 2^2}{3\times 2^5}$

$=\frac{2^{3+2}\times 3^4}{3\times 2^5}\left({\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right)$

$=\frac{2^5\times 3^4}{3\times 2^5}$

$2^{5-5}\times 3^{4-1}\left(a^m÷a^n=a^{m-n}\right)$

$2^0\times 3^3=1\times 3^3=3^3$

(ii) [(5²)³ × 5⁴] ÷ 5⁷

= [5^{2 × 3} × 5⁴] ÷ 5⁷ (a^m)ⁿ = a^mn

= [5⁶ × 5⁴] ÷ 5⁷

= [5^{6 + 4}] ÷ 5⁷ (a^m × aⁿ = a^m+n)

= 5¹⁰ ÷ 5⁷

= 5^{10 − 7} (a^m ÷ aⁿ = a^m−n)

= 5³

(iii) 25⁴ ÷ 5³ = (5 ×5)⁴ ÷ 5³

= (5²)⁴ ÷ 5³

= 5^{2 × 4} ÷ 5³ (a^m)ⁿ = a^mn

= 5⁸ ÷ 5³

= 5^{8 − 3} (a^m ÷ aⁿ = a^m−n)

= 5⁵

(iv) $\frac{3\times 7^2\times {11}^8}{21\times {11}^3}=\frac{3\times 7^2\times {11}^8}{3\times 7\times {11}^3}$

$3^{1-1}\times 7^{2-1}\times {11}^{8-3}\left(a^m÷a^n=a^{m-n}\right)$

$3^0\times 7^1\times {11}^5$ = 1 × 7 × 11⁵ = 7 × 11⁵

(v)  $\frac{3^7}{3^4\times 3^3}=\frac{3^7}{3^{4+3}}\left(a^m\times a^n=a^{m+n}\right)$

$=\frac{3^7}{3^7}=3^{7-7}$= $3^0\left(a^m÷a^n=a^{m-n}\right)$

= 1

(vi) 2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3

(vii) 2⁰ × 3⁰ × 4⁰ = 1 × 1 × 1 = 1

(viii) (3⁰ + 2⁰) × 5⁰ = (1 + 1) × 1 = 2

(ix) $\frac{2^8\times {\mathrm{a}}^5}{4^3\times {\mathrm{a}}^3}=\frac{2^8\times {\mathrm{a}}^5}{(2\times 2{)}^3\times {\mathrm{a}}^3}=\frac{2^8\times {\mathrm{a}}^5}{{\left(2^2\right)}^3\times {\mathrm{a}}^3}$

$=\frac{2^8\times {\mathrm{a}}^5}{\left(2^{2\times 3}\right)\times {\mathrm{a}}^3}$     (a^m)ⁿ = a^mn

= $\frac{2^8\times {\mathrm{a}}^5}{2^6\times {\mathrm{a}}^3}$

= $2^{8-6}\times a^{5-3}\left(a^m÷a^n=a^{m-n}\right)$

= $2^2\times {\mathrm{a}}^2=(2\times \mathrm{a}{)}^2$

= 2a²

(x) $\left(\frac{a^5}{a^3}\right)\times a^8=a^{5-3}\times a^8$

$=a^2\times a^8$

$=a^{2+8}=a^{10}$

(xi) $\frac{4^5\times a^8{b}^3}{4^5\times a^5{b}^2}=4^{5-5}\times a^{8-5}\times b^{3-2}$

$=4^0\times a^3\times b^1=1\times a^3\times b=a^{3}b$

(xii) (2³ × 2)² =  ^{${\left(2^{3+1}\right)}^2$}(a^m × aⁿ = a^m+n)

= (2⁴)² = 2^{4 × 2} (a^m)ⁿ = a^mn

= 2⁸

P3. Say true or false and justify your answer:

(i) 10 × 10¹¹ = 100¹¹

(ii) 2³ > 5²

(iii) 2³ × 3² = 6⁵

(iv) 3⁰ = (1000)⁰

Sol. (i) 10 × 10¹¹ = 100¹¹

L.H.S. = 10 × 10¹¹ = 10^{11 + 1} (a^m × aⁿ = a^m+n) = 10¹²

R.H.S. = 100¹¹ = (10 ×10)¹¹ = (10²)¹¹ = 10^{2 × 11} = 10²² (a^m)ⁿ = a^mn

As L.H.S. ≠ R.H.S.,

Therefore, the given statement is false.

(ii) 2³ > 5²

L.H.S. = 2³ = 2 × 2 × 2 = 8

R.H.S. = 5² = 5 × 5 = 25

As 25 > 8,

Therefore, the given statement is false.

(iii) 2³ × 3² = 6⁵

L.H.S. = 2³ × 3² = 2 × 2 × 2 × 3 × 3 = 72

R.H.S. = 6⁵ = 7776

As L.H.S. ≠ R.H.S.,

Therefore, the given statement is false.

(iv) 3⁰ = (1000)⁰

L.H.S. = 3⁰ = 1

R.H.S. = (1000)⁰ = 1 = L.H.S.

Therefore, the given statement is true.

P4. Express each of the following as a product of prime factors only in exponential form:

(i) 108 × 192

(ii) 270

(iii) 729 × 64

(iv) 768

Sol. (i) 108 × 192

= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)

= (2² × 3³) × (2⁶ × 3)

= 2^{6 + 2} × 3^{3 + 1}     (a^m × aⁿ = a^m+n)

= 2⁸ × 3⁴

(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3³ × 5

(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2) = 3⁶ × 2⁶

(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3

P5.Simplify:

(i) $\frac{{\left(2^5\right)}^{\mathbf{2}}\mathbf{\times }{\mathbf{7}}^{\mathbf{3}}}{{\mathbf{8}}^{\mathbf{3}}\mathbf{\times }\mathbf{7}}$

(ii) $\frac{\mathbf{25}\mathbf{\times }{\mathbf{5}}^{\mathbf{2}}\mathbf{\times }{\mathbf{t}}^{\mathbf{8}}}{{\mathbf{10}}^{\mathbf{3}}\mathbf{\times }{\mathbf{t}}^{\mathbf{4}}}$

(iii) $\frac{{\mathbf{3}}^{\mathbf{5}}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{\times }\mathbf{25}}{{\mathbf{5}}^{\mathbf{7}}\mathbf{\times }{\mathbf{6}}^{\mathbf{5}}}$

Sol. (i) $\frac{{\left(2^5\right)}^2\times 7^3}{8^3\times 7}=\frac{2^{3\times 2}\times 7^3}{(2\times 2\times 2{)}^3\times 7}$

$=\frac{2^{10}\times 7^3}{2^{3\times 3}\times 7}$

= $2^{10-9}\times 7^{3-1}=2^1\times 7^2$ = 98

(ii) $\frac{25\times 5^2\times t^8}{{10}^3\times t^4}=\frac{5\times 5\times 5^2\times t^8}{(5\times 2{)}^3\times t^4}$

=$\frac{5^{1+1+2}\times t^8}{5^3\times 2^3\times t^4}=\frac{5^4\times t^8}{5^3\times 2^3\times t^4}$

$=\frac{5^{4-3}\times t^{8-4}}{2^3}$

$=\frac{5\times t^4}{2\times 2\times 2}=\frac{5t^4}{8}$

(iii) $\frac{3^5\times {10}^5\times 25}{5^7\times 6^5}=\frac{3^5\times (2\times 5{)}^5\times 5\times 5}{5^7\times 2^5\times 3^5}$

$=\frac{3^5\times 2^5\times 5^5\times 5^2}{5^7\times 2^5\times 3^5}=\frac{3^5\times 2^5\times 5^7}{5^7\times 2^5\times 3^5}$

= $3^{5-5}\times 2^{5-5}\times 5^{7-7}=3^0\times 5^0\times 7^0$

= 1× 1 × 1 = 1

EXERCISE-12.3

P1.Write the following numbers in the expanded forms:

279404, 3006194, 2806196, 120719, 20068

Sol. 279404 = 2 × 10⁵ + 7 × 10⁴ + 9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰

3006194 = 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰

2806196 = 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰

120719 = 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰

20068 = 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

P2. Find the number from each of the following expanded forms:

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰

(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰

(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹

Sol. (a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰ = 86045

(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰ = 405302

(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰ = 30705

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹ = 900230

P3. Express the following numbers in standard form:

(i) 5, 00, 00, 000                (ii) 70, 00, 000

(iii) 3, 18, 65, 00, 000       (iv) 3, 90, 878

(v) 39087.8                           (vi) 3908.78

Sol. (i) 50000000 = 5 × 10⁷

(ii) 7000000 = 7 × 10⁶

(iii) 3186500000 = 3.1865 × 10⁹

(iv) 390878 = 3.90878 × 10⁵

(v) 39087.8 = 3.90878 × 10⁴

(vi) 3908.78 = 3.90878 × 10³

P4. Express the number appearing in the following statements in standard form.

(a) The distance between Earth and Moon is 384, 000, 000 m.

(b) Speed of light in vacuum is 300, 000, 000 m/s.

(c) Diameter of the Earth is 1, 27, 56, 000 m

(d) Diameter of the Sun is 1, 400, 000, 000 m

(e) In a galaxy there are on an average 100, 000, 000, 000 stars.

(f) The universe is estimated to be about 12, 000, 000, 000 years old.

(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300, 000, 000, 000, 000, 000, 000 m

(h) 60, 230, 000, 000, 000, 000, 000, 000 molecules are contained in a drop of water weighing 1.8 gm.

(i) The earth has 1, 353, 000, 000 cubic km of sea water.

(j) The population of India was about 1, 027, 000, 000 in March, 2001.

Sol. (a) 3.84 × 10⁸ m

(b) 3 × 10⁸ m/s

(c) 1.2756 × 10⁷ m

(d) 1.4 × 10⁹ m

(e) 1 × 10¹¹ stars

(f) 1.2 × 10¹⁰ years

(g) 3 × 10²⁰ m

(h) 6.023 × 10²²

(i) 1.353 × 10⁹ cubic km

(j)1.027 × 10⁹