Exponents And Powers

(i) In the given question,

We have to find the value of 2⁶

2⁶ = 2 × 2 × 2 × 2 × 2 × 2

= 64

(ii) In the given question,

We have to find the value of 9³

9³ = 9 × 9 × 9

= 729

(iii) In the given question,

We have to find the value of 11²

11² = 11 × 11

= 121

(iv) In the given question,

We have to find the value of 5⁴

5⁴ = 5 × 5 × 5 × 5

= 625

(i) We have,

6 × 6 × 6 × 6

= 6^1+1+1+1

= 6⁴

(ii) We have,

t × t

= t¹⁺¹

= t²

(iii) We have,

b × b × b × b

= b^1+1+1+1

= b⁴

(iv) We have,

5 × 5 × 7 × 7 × 7

= 5¹⁺¹ × 7¹⁺¹⁺¹

= 5² × 7³

(v) We have,

2 × 2 × a × a

= 2¹⁺¹ × a¹⁺¹

= 2² × a²

(vi) We have,

a × a × a × c × c × c × c × d

= a¹⁺¹⁺¹ × c^1+1+1+1 × d¹

= a³ × c⁴ × d

(i) 512

We have,

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2⁹

(ii) 343

We have,

343 = 7 × 7 × 7

= 7³

(iii) 729

We have,

729 = 3 × 3 × 3 × 3 × 3 × 3

= 3⁶

(iv) 3125




We have,

3125 = 5 × 5 × 5 × 5 × 5

= 5⁵

(i) We have, 4³ or 3⁴

On simplifying we get,

4³ = 4 × 4 × 4

= 64

And,

3⁴ = 3 × 3 × 3 × 3

= 81

Clearly,

81 > 64

Thus,

3⁴ > 4³

(ii) We have, 5³ or 3⁵

On simplifying we get,

5³ = 5 × 5 × 5

= 125

3⁵ = 3 × 3 × 3 × 3 × 3

= 243

Clearly,

243 > 125

Thus,

3⁵ > 5³

(iii) We have, 2⁸ or 8²

On simplifying we get,

2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 256

8² = 8 × 8

= 64

Clearly,

256 > 64

Thus,

2⁸ > 8²

(iv) We have, 100² or 2¹⁰⁰

On simplifying we get,

100² = 100 × 100

= 10000

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

= 1024

Now,

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

Clearly,

2¹⁰⁰ > 10000

Thus,

2¹⁰⁰ > 100²

(v) W have, 2¹⁰ or 10²

On simplifying we get,

10² = 10 × 10

= 100

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

= 1024

Clearly,

1024 > 100

Thus,

2¹⁰ > 10²

(i) On simplifying we get,

648 = 2 × 2 × 2 × 3 × 3 × 3 × 3

= 2³ × 3⁴

(ii) On simplifying we get,

405 = 3 × 3 × 3 × 3 × 5

= 3⁴ × 5

(iii) On simplifying we get,

540 = 2 × 2 × 3 × 3 × 3 × 5

= 2² × 3³ × 5

(iv) On simplifying we get,

3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= 2⁴ × 3² × 5²

(i) We have,

2 × 10³ = 2 × 10 × 10 × 10

= 2 × 1000

= 2000

(ii) We have,

7² × 2² = 7 × 7 × 2 × 2

= 49 × 4

= 196

(iii) We have,

2³ × 5 = 2 × 2 × 2 × 5

= 8 × 5

= 40

(iv) We have,

3 × 4⁴ = 3 × 4 × 4 × 4 × 4

= 3 × 256

= 768

(v) We have,

0 × 10² = 0 × 10 × 10

= 0 × 100

= 0

(vi) We have,

5² × 3³ = 5 × 5 × 3 × 3 × 3

= 25 × 27

= 675

(vii) We have,

2⁴ × 3² = 2 × 2 × 2 × 2 × 3 × 3

= 16 × 9

= 144

(viii) We have,

3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10

= 9 × 10000

= 90000

(i) We simplify the given expression (-4)³ as:

(-4)³ = (-4) × (-4) × (-4)

= - 64

(ii) We simplify the given expression (-3) × (-2)³ as:

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

= (-3) × (-8)

= 24

(iii) We simplify the given expression (-3)² × (-5)² as:

(-3)² × (-5)² = (-3) × (-3) × (-5) × (-5)

= 9 × 25

= 225

(iv) We simplify the given expression (-2)³ × (-10)³ as:

(-2)³ × (-10)³ = (-2) × (-2) × (-2) × (-10) × (-10) × (-10)

= (-8) × (-1000)

= 8000

(i) We have,

2.7 × 10¹² and 1.5 × 10⁸

Since,

2.7 is greater than 1.5 and 10¹² is also greater than 10⁸

Hence,

2.7 × 10¹² > 1.5 × 10⁸

(ii) We have,

4 × 10¹⁴ and 3 × 10¹⁷

Since,

4 is greater than 3 but 10¹⁷ is greater than 10¹⁴

Hence,

3 × 10¹⁷ > 4 × 10¹⁴

(i) We know that, (a^m × aⁿ = a^{m + n})

Thus,

3² × 3⁴ × 3⁸

= (3)^{2 + 4 + 8}

= 3¹⁴

(ii) We have,

6¹⁵ 6¹⁰

We know that,

(a^m aⁿ = a^{m - n})

Thus,

6¹⁵ 6¹⁰

= (6)^{15 - 10}

= 6⁵

(iii) We have,

a³ × a²

We know that,

(a^m × aⁿ = a^{m + n})

Therefore,

a³ × a²

= (a)^{3 + 2}

= a⁵

(iv) We have,

7^x × 7²

We know that,

(a^m × aⁿ = a^{m + n})

Thus,

7^x × 7²

= (7)^{x + 2}

(v) We have,

(5²)³ 5³

Using identity:

(a^m)ⁿ = a^{m × n}

= 5^{2 × 3} 5³

= 5⁶ 5³

We know that,

(a^m aⁿ = a^{m - n})

Thus,

5⁶ 5³

= (5)^{6 - 3}

= 5³

(vi) We have,

2⁵ × 5⁵

We know that,

[a^m × b^m = (a × b)^m]

Thus,

2⁵ × 5⁵

= (2 × 5)^{5 + 5}

= 10⁵

(vii) We have,

a⁴ × b⁴

We know that,

[a^m × b^m = (a × b)^m]

Thus,

a⁴ × b⁴

= (a × b)⁴

(viii) We have,

(3⁴)³

We know that,

(a^m)ⁿ = a^mn)

Thus,

(3⁴)³

= (3⁴)³

= 3¹²

(ix) We have,

(2²⁰ 2¹⁵) × 2³

We know that,

(a^m aⁿ = a^{m - n})

Thus,

(2^{20 - 15}) × 2³

= (2)⁵ × 2³

We know that,

(a^m × aⁿ = a^{m + n})

Thus,

(2)⁵ × 2³

= (2^{5 + 3})

= 2⁸

(x) We have,

(8^t 8²)

We know that,

(a^m aⁿ = a^{m - n})

Thus,

(8^t 8²)

= (8^{t – 2})

(i) In the above question,

We have to simplify the given numbers into exponential form:

∴ We have,

=

=

=

= (a^m × aⁿ = a^{m + n})

Using identity: (a^m aⁿ = a^{m - n})

= 2^{5 - 5} × 3^{4 – 1}

= 2⁰3³

= 1 × 3³

= 3³

(ii) In the above question,

We have to simplify the given numbers into exponential form:

[(5²)³ × 5⁴] 5⁷

Using identity: (a^m)ⁿ = a^mn)

= [(5)^{2 × 3} × 5⁴] 5⁷

= [(5)⁶ × 4] 5⁷

Using identity: (a^m × aⁿ = a^{m + n})

= [5^{6 + 4}] 5⁷

Using identity: (a^m aⁿ = a^{m - n})

Therefore,

= 5¹⁰ 5⁷

= 5^{10 – 7}

= 5³

(iii) In the above question,

We have to simplify the given numbers into exponential form:

We have,

25⁴ 5³

= (5 × 5)⁴ 5³

Using identity: (a^m)ⁿ = a^mn

= 5^{2 × 4} 5³

= 5⁸ 5³

Using identity: (a^m aⁿ = a^{m - n})

5⁸ 5³

= 5^{8 – 3}

= 5⁵

(iv) In the above question,

We have to simplify the given numbers into exponential form:

We have,

=

Using identity: (a^m aⁿ = a^{m - n})

= 3^{1 - 1} × 7^{2 – 1} × 11^{8 – 3}

= 3⁰ × 7¹ × 11⁵

= 1 × 7 × 11⁵

= 7 × 11⁵

(v) In the above question,

We have to simplify the given numbers into exponential form:

We have,

Using identity: (a^m × aⁿ = a^{m + n})

=

Using identity (a^m aⁿ = a^{m - n})

= 3^{7 - 7}

= 3^o

= 1

(vi) In the above question,

We have to simplify the given numbers into exponential form:

∴ We have,

2⁰ + 3⁰ + 4⁰

= 1 + 1 + 1

= 3

(vii) In the above question,

We have to simplify the given numbers into exponential form:

∴ We have,

2⁰ × 3⁰ × 4⁰

= 1 × 1 × 1

= 1

(viii) In the above question,

We have to simplify the given numbers into exponential form:

∴ We have,

(3⁰ + 2⁰) × 5⁰

= (1 + 1) × 1

= 2

(ix) In the above question,

We have to simplify the given numbers into exponential form:

∴ We have,

=

Using identity: (a^m)ⁿ = a^mn

Using identity: (a^m aⁿ = a^{m - n})

= 2^{8 – 6} × a^{5 - 3}

= 2² × a²

Using identity [a^m ×b^m = (a × b)^m]

= (2 × a)²

= (2a)²

(x) In the above question,

We have to simplify the given numbers into exponential form:

∴ We have,

() × a⁸

Using identity: (a^m aⁿ = a^{m - n})

= a^{5 – 3} × a⁸

= a² × a⁸

Using identity (a^m × aⁿ = a^{m + n})

= a^{2 + 8}

= a¹⁰

(xi) In the above question,

We have to simplify the given numbers into exponential form:

∴ We have,

Using identity: (a^m aⁿ = a^{m - n})

= 4^{5 – 5} × a^{8 – 5} × b^{3 – 2}

= 4⁰ × a³ × b¹

= 1 × a³ × b

= a³b

(xii) In the above question,

We have to simplify the given numbers into exponential form:

∴ We have,

(2³ × 2)²

Using identity: (a^m × aⁿ = a^{m + n})

= (2^{3 + 1})²

= (2⁴)²

Using identity: (a^m)ⁿ = a^mn

Therefore,

= 2^{4 × 2}

= 2⁸

(i) We have,

10 × 10¹¹ = 100¹¹

L.H.S. = 10 × 10¹¹

Using identity: (a^m × aⁿ = a^{m + n})

= 10^{11 + 1}

= 10¹²

R.H.S. = 100¹¹

= (10 × 10)¹¹

= (10²)¹¹

Using identity: (a^m)ⁿ = a^mn

= 10^{2 × 11}

= 10²²

So, it is clear that

L.H.S. R.H.S.

∴ The statement given in the question is false

(ii) We have,

2³ > 5²

L.H.S. = 2³

= 2 × 2 × 2

= 8

R.H.S. = 5²

= 5 × 5

= 25

Since,

It is clear that 25 > 8

∴ The statement given in the question is false

(iii) We have,

2³ × 3² = 6⁵

L.H.S. = 2³ × 3²

= 2 × 2 × 2 × 3 × 3

= 72

R.H.S. = 6⁵

= 6 × 6 × 6 × 6 × 6

= 7776

It is clear that

L.H.S. R.H.S.

∴ The statement given in the question is false

(iv) We have,

3⁰ = (1000)⁰

L.H.S. = 3⁰

= 1

R.H.S. = (1000)⁰

= 1

Hence,

L.H.S. = R.H.S.

∴ The statement given in the question is true

(i) We have,

108 × 192

We have to express this as a product of prime factors only in exponential form

∴ 108 × 192

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

= (2² × 3³) × (2⁶ × 3)

Using identity: (a^m × aⁿ = a^{m + n})

= 2^{6 + 2} × 3^{3 + 1}

= 2⁸ × 3⁴

(ii) We have,

270

We have to express this as a product of prime factors only in exponential form

Thus,

270

= 2 × 3 × 3 × 3 × 5

= 2 × 3³ × 5

(iii) We have,

729 × 64

We have to express this as a product of prime factors only in exponential form

Thus,

729 × 64

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

= 3⁶ × 2⁶

(iv) We have,

768

We have to express this as a product of prime factors only in exponential form

Thus, 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2⁸ × 3

(i)

We have,

Using identity: (a^m)ⁿ = a^mn , we get,

Using identity: (a^m)ⁿ = a^mn, we get,

=

Using identity: (a^m aⁿ = a^{m - n})

= 2^{10 - 9} × 7^{3 – 1}

= 2¹ × 7²

= 2 × 7 × 7

= 98

(ii) We have,

Using identity: [(a × b)^m = a^m ×b^m]

Using identity: (a^m × aⁿ = a^{m + n})

=

By using (a^m aⁿ = a^{m - n}), we get:

=

=

(iii) We have,

=

Using identity: [(a × b)^m = a^m ×b^m]

=

Using identity: (a^m × aⁿ = a^{m + n})

By using (a^m aⁿ = a^{m - n}), we get:

= 3^{5 – 5} × 2^{5 – 5} × 5^{7 – 7}

= 3⁰ × 2⁰ × 5⁰

= 1 × 1 × 1

= 1

In this question,

We have to write the above given numbers into expanded form:

∴ 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

(a) In this question,

We have to find the number from above given numbers which are in expanded form:

Thus,

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

= 80000 + 6000 + 00 + 40 + 5

= 86045

(b) In this question,

We have to find the number from above given numbers which are in expanded form:

Thus,

4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰

= 400000 + 5000 + 300 + 2

= 405302

(c) In this question,

We have to find the number from above given numbers which are in expanded form:

Thus,

3 × 10⁴ + 7 × 10² + 5 × 10⁰

= 30000 + 700 + 5

= 30705

(d) In this question,

We have to find the number from above given numbers which are in expanded form:

Thus,

9 × 10⁵ + 2 × 10² + 3 × 10¹

= 900000 + 200 + 30

= 900230

(i) We have to express the given number in standard form:

Thus,

5, 00, 00, 000 = 5 × 10⁷

(ii) We have to express the given number in standard form:

Thus,

70, 00, 000 = 7 × 10⁶

(iii) We have to express the given number in standard form:

Thus,

3, 18, 65, 00, 000 = 3.1865 × 10⁹

(iv) We have to express the given number in standard form:

Thus,

3, 90, 878 = 3.90878 × 10⁵

(v) We have to express the given number in standard form:

Thus,

39087.8 = 3.90878 × 10⁴

(vi) We have to express the given number into the standard form:

Thus,

3908.78 = 3.90879 × 10³

(a) It is given that,

Distance between Earth and Moon = 384, 000, 000 m

Thus,

384, 000, 000 = 3.84 × 10⁸ m

(b) It is given that,

Speed of light in vacuum = 300, 000, 000 m/s

Thus,

300, 000, 000 = 3 × 10⁸ m/s

(c)It is given that,

Diameter of Earth = 1, 27, 56, 000 m

Thus,

1, 27, 56, 000 = 1.2756 × 10⁷ m

(d) It is given that,

Diameter of Sun = 1, 400, 000, 000 m

Thus,

1, 400, 000, 000 = 1.4 × 10⁹ m

(e) It is given that,

Average number of stars in galaxy = 100, 000, 000, 000 stars

Thus,

100, 000, 000, 000 = 1 × 10¹¹ stars

(f) It is given that,

Estimated age of universal = 12, 000, 000, 000 years

Thus,

12, 000, 000, 000 = 1.2 × 10¹⁰ years

(g) It is given that,

Distance of sun from the centre of Milky Way galaxy = 300, 000, 000, 000, 000, 000, 000 m

Thus,

300, 000, 000, 000, 000, 000, 000 = 3 × 10²⁰ m

(h) It is given that,

Total number of molecules in a drop of water weighing 1.8 gram = 60, 230, 000, 000, 000, 000, 000, 000 m

Thus,

60, 230, 000, 000, 000, 000, 000, 000 = 6.023 × 10²²

(i) It is given that,

Area of sea water on earth = 1, 353, 000, 000 cubic km

Thus,

1, 353, 000, 000 = 1.353 × 10⁹ cubic km

(j) It is given that,

Population of India in March 2001= 1, 027, 000, 000

Thus,

1, 027, 000, 000 = 1.027 × 10⁹