Exponents And Powers

Here, 2 is rational number (constant) as base in expression and n is the power of that base 2.

n is called as called exponent (power or index) in the given expression.

Hence, the correct option is C.

For the fixed base, if the exponent decrease by 1, the number becomes one-tenth of the previous number.

e.g; for 10⁷, when the exponent is decreased by 1

it becomes 10^7-1 = 10⁶

Now,

Hence option A is the correct option.

As per the law of exponent

,

where a is non-zero integer

So

Hence the correct option is B.

According to the law of exponent

where a is non zero integer

⇒

Hence, the correct option is A.

According to the rule of exponent

The value of

Hence, the correct option is C

As per the law of the exponent

, where a is non-zero integer

⇒ Value of

Hence, the correct option C.

As per the law of exponent

where a is non zero integer

⇒ The reciprocal of

Hence, the correct option is B

As per the law of exponent

where a is non zero integer

If a is any rational number then its multiplicative inverse is given by

Such that

⇒ The multiplicative inverse of

Hence the correct option is C

Given (-2)^{2× 3-1}

Applying Bodmas rule

Multiplication operation will be carried before subtraction

(-2)^{2× 3 -1} = (-2)^6-1 = (-2)⁵

-2 × -2 × -2 × -2 × -2 = -32

(for -a^m , if m is odd then (–a)^m is some negative integer.)

Hence the correct option is C

The value of

For -a^m if m is even, (-a)^m is positive

The multiplicative inverse of ‘a’ is given by where

The multiplicative inverse is

Hence, the correct option is C

Using the laws of exponent

a^m×aⁿ = a^{(m + n)} where a is any non zero integer

Hence the correct option is B

As per the law of exponent

a⁰ = 1 (where a is non-zero integer)

Similarly, y⁰ = 1

Hence, the correct option is A.

as per the law of exponent

Similarly,

Hence the correct option is B.

Using the law of exponent

Similarly,

Hence the correct option is C

As per the laws of exponent

(a^m)ⁿ = a^{(m× n)}where a is non-zero integer

Similarly, (x^m)ⁿ = x ^{m× n}

Hence the correct option is B

using laws of exponent

where a is non zero integer

Similarly, (

Hence, the correct option is D.

using law of exponent

where a is non zero integer

Similarly

Hence the correct option is D.

using law of exponent

Similarly,

Hence the correct option is B

using law of exponent

Hence the correct option is D

By laws of exponent,

x^m ÷ xⁿ = x^m-n

In this question,

m = 7

n = 12

∴x^m ÷ xⁿ = x^m-n

⇒ x⁷ ÷ x¹² = x^7-12

⇒ x⁷ ÷ x¹² = x^-5

Hence, for a non-zero integer x, x⁷ ÷ x¹² is equal to x^-5.

By laws of exponent,

(x^m)ⁿ = x^mn

In this question,

m = 4

n = -3

∴ (x^m)ⁿ = x^mn

⇒ (x⁴)^-3 = x^4×-3

⇒ (x⁴)^-3 = x^-12

Hence, for a non-zero integer x, (x⁴)^-3 is equal to x^-12.

By laws of exponent,

∴ By using,

Again using,

= 56 – 12= 44

Hence, the value of (7^–1 – 8^–1)^–1 – (3^–1 – 4^–1)^–1 is 44.

Standard form is a form of writing very large or very small numbers without negative exponent and in a précised form, with decimal after one digit.

We know that,

By laws of exponent,

⇒ 0.000064 = 6.4 × 10^-5

Hence, the standard form for 0.000064 is 6.4 × 10^-5.

Standard form is a form of writing very large or very small numbers without negative exponent and in a précised form, with decimal after one digit.

We know that,

234000000 = 234 × 1000000

234000000 = 2.34 × 100 × 10⁶

234000000 = 2.34 × 10² × 10⁶

By laws of exponent,

x^m × xⁿ = x^{m + n}

⇒234000000 = 2.34 × 10^{2 + 6}

⇒ 234000000 = 2.34 × 10⁸

Hence, the standard form for 234000000 is 2.34 × 10⁸.

By laws of exponent,

⇒ 2.03 × 10^-5 = 0.0000203

Hence, the usual form for 2.03 × 10^–5 is 0.0000203.

By laws of exponent,

a⁰ = 1

where, a≠ 1

Hence, is equal to 1.

By laws of exponent,

x^m ÷ y^m = (x ÷ y)^m

Hence, is equal to .

By laws of exponent,

x^m ÷ y^m = (x ÷ y)^m

⇒ x⁴ ÷ y⁴ = (x ÷ y)⁴

Hence, for any two non-zerorational numbers x and y, x⁴ ÷ y⁴ is equal to (x ÷ y)⁴

By laws of exponent,

x^m ÷ xⁿ = x^m-n

In this question,

m = 13

n = 8

∴ x^m ÷ xⁿ = x^m-n

⇒ p¹³ ÷ p⁸ = p^13-8

⇒ p¹³ ÷ p⁸ = p⁵

Hence, for a non-zero rational number p, p¹³ ÷ p⁸ is equal to p⁵.

By laws of exponent,

(x^m)ⁿ = x^mn

In this question,

m = -2

n = 3

∴ (x^m)ⁿ = x^mn

⇒ (z^-2)³ = z^-2×3

⇒ (z^-2)³ = z^-6

Hence, for a non-zero rational number z, (z^-2)³ is equal to z^-6.

Cube of x = x × x × x

Hence, Cube of -1/2 is

Hence, is not the reciprocal of .

multiplicative is inverse or reciprocal of number.

⇒ Multiplicative inverse of is .

⇒ So, the multiplicative inverse of 10¹⁰ will

∴ Multiplicative inverse of 10¹⁰ =10^-10

Hence, multiplicative factor of 10¹⁰ is 10^–10

We know the formula –

⇒ a^m× aⁿ = a^m+n

⇒ a³× a^-10 =a^{3 – 10}

∴ a³× a^-10 =a^{– 7}

As we know that,

⇒ a⁰ = 1

∴ 5⁰ = 1

we know that,

a ^{– b} =

∴ 5⁵× 5 ^{– 5} = 5⁵×

⇒ 5⁵× 5 ^{– 5} = 1

⇒

⇒

⇒

In 8^–2, 8 is the base and – 2 is the power.

⇒ 8^–2 =

∴ 8^–2 = (Here, base is 64 and power is 1)

We know that,

2⁶ = 2×2×2×2×2×2 = 64

⇒ Base is 2 and power is 6.

∴ (Base is 2 and power is –6)

Very small numbers can be expressed in standard form by using negative exponents.

For example – 0.0045 can be written as 4.5× 10^–3

Very large numbers can be expressed standard form by using positive exponents.

For example – 45,000 can be expressed as 45× 10³

We know that,

a^m× aⁿ = a^m+n

⇒ (10)⁵× (10)^-10 = (10)^5–10

∴ (10)⁵× (10)^-10 = (10)^–5

We know that,

a^m ÷ aⁿ = a^m–n

a^m× aⁿ = a^m+n

⇒ =

⇒

⇒

⇒

we know that,

a^-m =

⇒ [4^–1 +3^–1 + 6^–2]^–1 =

⇒ [4^–1 +3^–1 + 6^–2]^–1 =

⇒ [4^–1 +3^–1 + 6^–2]^–1 = {∵ L.C.M. of (4,3,36) is 36}

⇒ [4^–1 +3^–1 + 6^–2]^–1 =

Hence, [4^–1 +3^–1 + 6^–2]^–1 = .

We know that,

a⁰ = 1

[2^–1 + 3^–1 + 4^–1]⁰ = 1

Given form is .

⇒

⇒

∴ Standard form of is 1× 10^-8.

Standard form of 12340000 = 1234× 10⁴

⇒ 12340000 = 1.234× 10⁴ × 10³

Hence, standard form of 12340000 is 1.234× 10⁷.

3.41 × 10⁶ = 3.41× 10×10×10×10×10×10

⇒ 3.41 × 10⁶ =3410000

Hence. usual form of 3.41 × 10⁶ is 3410000.

2.39461 × 10⁶ = 2.39461× 10×10×10×10×10×10.

⇒ 2.39461 × 10⁶=2394610

Hence, usual form of 2.39461 × 10⁶ is 2394610.

we know that,

a^-m =

⇒ 36 = 6× 6

.

Hence, expressed as a power with the base 6 is 6 ^–2

⇒

equal

⇒ 325 × 100000000 = 325 × 10⁸

⇒

11

= (6⁰ – 7⁰) × (6⁰ + 7⁰)

= (0-0) × (0 + 0)

= 0

= 36 + 9 + 4 = 49

False

LHS = (-4)^-2 =

RHS = (4)^-2 =

LHS equal to RHS

Therefore these numbers are same and not multiplicative inverse

True

⇒ Not equal to 1

∴ They are not multiplicative inverse

True

False

True

⇒329.25

= 3 × 10² + 2 × 10¹ + 9 × 1 + 2 × 10^–1 + 5 × 10^–2

False

False

False

False

⇒5⁰ = 1

False

⇒ (-2)⁰ = 1

False

⇒ ()⁰ = 1

False

⇒ (-6)⁰ = 1

True

⇒ (-7)^{-4 + 2} = (-7)^-2

True

False

the above statement is false.

here , from the laws of exponents , (ab)^m= a^m × b^m

So , here , (ab)^pq = a^pq × b^pq ≠ a^p × b^q

∴ LHS ≠ RHS

∴ a^p × b^q ≠ (ab)^pq

The above statement is true.

using law of exponents , =

Also , =

So , we can say that , = =

Hence , LHS = RHS.

the above statement is true.

as we know , = a^-m , so here , we have = a^-(-m) = a^m

Hence , LHS = RHS

the above statement is true.

→ (–2)⁴ × = (2)⁴ × = (2)⁴ × = = 5⁴ …..{using laws of exponents ie. = }

the above statement is true.

As we know that, from the standard form ,

0.000037 = = = 3.7 × 10^–5

the above statement is true.

From the standard form , 203000 = 203 × = 2.03 × × = 2.03 × 10⁵

the above statement is false.

We know that , 2 × 10^-2= 2 × = 2 × = = 0.02

the above statement is false.

Because , 5^-2= = ≠ 25. …….(using laws of exponents ie.a^-m=

the above statement is true.

For example , 203000 = 203 × 10 × 10 × 10 = 203 × 10³ = 2.03 × 10² × 10³ = 2.03 × 10⁵

As we can see above , Large numbers can be expressed in the standard form by using positive exponents.

the above statement is true.

LHS = (ab)^m=(a×b)^m = a^m × b^m ……….(by law of exponents)

as we know 100² = 100 × 100 = 10000

100¹ = = 1000

100⁰ = = 1

Continuing the above pattern , we get ,

100^-1 =

Similarly , 100^-2= ÷ 100 = × = =

100^-3 = ÷ 100 = ÷ 100 = =

Similarly we can say that , 100^-10=

we know that, 2^-2=

And 2^-3 =

From the law of exponents , we know that a^m × aⁿ = a^(m+n)

∴ 2^-2 × 2^-3 = × = = = = 2^-5

∴ 2^-2 × 2^-3 = 2^-5

by laws of exponents , we know that , a^m÷ aⁿ = a^(m-n)

So , ÷ = = = =

∴ ÷ =

using law of exponents , =

Also , =

So , 3^–5 × 3^–4 = 3^(-5)+(-4) = 3^-9=

∴ 3^–5 × 3^–4 =

as we know that , 2 × 2 × 2 × 2 = 16

∴ 16 = 2⁴

∴ 16^-2 =(2⁴)^-2=2^-8…..(by the laws of exponents ie. (a^m)ⁿ = a^(mn)

we know that , 27 = 3 × 3 × 3 = 3³

And , -27 = -3 × -3 × -3 = (-3)³

And , 64 = 4 × 4 × 4 = 4³

∴ = =

∴ = = …..(by laws of exponents ie. = )

we know that , 16 = 4 × 4

And 81 = 9 × 9

∴ = =

∴ = = - …..(by laws of exponents ie. = )

using laws of exponents , (a^m)ⁿ = a^(mn)

And , =

We have , = = =

∴ =

using laws of exponents ,a^m × aⁿ = a^(m+n)

And , a^m ÷ aⁿ = a^(m-n)

We have , (2⁵ ÷ 2⁸) = 2^(5-8)= 2^-3

Now , 2^-3 × 2^–7= 2^(-3)+(-7)= 2^-10

∴ (2⁵ ÷ 2⁸) × 2^–7= 2^-10

cube of -2 = (-2)³ = -(2)³

And square of +4 = 4²

∴ the product of the cube of (–2) and the square of (+4)

= -(2)³ × 4²= -8 × 16 = -128

∴ the product of the cube of (–2) and the square of (+4) is -128.

from the laws of exponents , we know that ,

=

So, we have + + = 4² + 2² + 3² = 16 + 4 + 9 = 29

∴ + + = 29

from the laws of exponents , we know that ,

=

And , a^m × aⁿ = a^(m+n)

And , a^m ÷ aⁿ = a^(m-n)

And , (a^m)ⁿ = a^(mn)

So , we have × × 3^-1 × = × 3⁴ × ×

= × 3⁴ × ×

= × 3⁴ ×

= ×3⁴×

=

=

=

=

=

∴ × × 3^-1 × =

from the laws of exponents , we know that ,

a^m ÷ aⁿ = a^(m-n)

So , we have =

=

=

=

=

∴ =

from the laws of exponents , we know that ,

a^m × aⁿ = a^(m+n)

And , a^m ÷ aⁿ = a^(m-n)

So ,we have , (2⁵ ÷ 2⁸) = 2^(5-8)= 2^-3

Now , 2^-3 × 2^–7= 2^(-3)+(-7)= 2^-10

∴ (2⁵ ÷ 2⁸) × 2^–7= 2^-10 = =

∴ (2⁵ ÷ 2⁸) × 2^–7 =

from the laws of exponents , we know that ,

a^m × aⁿ = a^(m+n)

So , we have , × = =

Now , =

On comparing both the sides , we get ,

-16 = 8x

x = = -2

∴ x = -2

from the laws of exponents , we know that,

a^m × aⁿ = a^(m+n)

So , we have , (–2)³ × (–2)^–6 =(-2)^(3)+(-6) = (-2)^(-3)

Now , we have , (-2)^(-3)= (-2)^(2x-1)

On comparing both the sides , we get ,

-3 = 2x-1

2x = -3 + 1

2x = -2

x = = -1

∴ x = -1

from the laws of exponents , we have ,

So , we have , (2¹ + 4¹ + 6¹ +8¹)^x =

=

=

Now , we have , = 1

As we know , a⁰ = 1

So , = 1 can be only possible when x=0.

∴ x = 0.

Given that to divide 293 by 10,00,000

10,00,000 can be written as 10⁶

∴ 10,00,000 = 10⁶

^⇒

Using the law of exponent,

^⇒

∴ The result in standard form is

Given that x = (100)^1–4 ÷ (100)⁰

Using the law of exponent,

^⇒

∴

To the value of x^–3

^⇒

Using the law of exponent,

^⇒

∴ The value of is

Let as assume the number is x.

Let x be multiply with (–29)⁰ and the product becomes ( + 29)⁰

So,

Using the law of exponent,

⇒ x × 1 = 1

∴ x = 1

∴The number is 1

Let as assume the number is x.

Let (–15)^–1 is divided by x to get the quotient (–15)^–1

So,

⇒

Using the law of exponent,

⇒

⇒

Using the law of exponent,

⇒

∴The number is 1

Remember, a is called multiplicative inverse of b, if a × b = 1.

∴ (–7)^–2 ÷ (90)^–1 =

=

=

∵ (-a)^m = a^m , if m is an even number and (-a)^m =

Put b = in a × b = 1

a × = 1

∴

∴ The multiplicative inverse of (–7)^–2 ÷ (90)^–1 is

Given that, 5^3x–1 ÷ 25 = 125

∵ 25 = 5 × 5 = 5² and 25 = 5 × 5 × 5 = 5³

∴ 5^3x–1 ÷ 5² = 5³

Using the law of exponent,

⇒ 5^3x-1-2 = 5³

⇒ 5^3x-3 = 5³

Comparing both sides,

3x-3 = 3

⇒3x = 6

⇒x = 2

∴ the value of x is 2

39,00,00,000 can be written as

39,00,00,000 = 39 × 10 × 10 × 10 × 10 × 10 × 10 × 10

= 39 × 10⁷

= 3.9 × 10¹ × 10⁷

Using the law of exponent,

⇒ 39,00,00,000 = 3.9 × 10⁸

∴ The standard form of 39,00,00,000 is 3.9 × 10⁸

0.000005678 can be written as

0.000005678 = 0.5678 × 10^-5

= 5.678 × 10^-1 × 10^-5

Using the law of exponent,

⇒ 0.000005678 = 5.678 × 10^-6

∴ The standard form of 0.000005678 is 5.678 × 10^-6

Given that, to find the product of 3.2 × 10⁶ and 4.1 × 10^–1

Product of 3.2 × 10⁶ and 4.1 × 10^–1 = (3.2 × 10⁶)( 4.1 × 10^–1)

= (3.2 × 4.1) × 10⁶ × 10^-1

Using the law of exponent,

⇒ 13.12 × 10⁵

= 1.312 × 10⁵ × 10¹

= 1.312 × 10⁶

∴ The product of 3.2 × 10⁶ and 4.1 × 10^–1 in the standard form is 1.312 × 10⁶.

Given that,

Using the law of exponent,

=

Using the law of exponent,

∴

The distance travelled by birds = 15,000 km

∵ 1 km = 1000 m

∴ 15,000 km = 15000 × 1000 m

= 15000000 m

= 15 × 10⁶ m

= 1.5 × 10⁷ m

∴ The distance in metres is 1.5 × 10⁷ m

The distance from the sun to Pluto = 59,1,30,00,000 m

∴ standard form of 59,1,30,00,000 = 5913 × 10⁶

= 5.913 × 10³ × 10⁶

Using the law of exponent,

= 5.913 × 10⁹

∴ The distance from the sun to Pluto is 5.913 × 10⁹

Weight = 0.00000001 gram

∴ standard form of 0.00000001 gram = 0.1 × 10^-7 g

= 1 × 10^-1 × 10^-7 g

Using the law of exponent,

= 1 × 10^-8 g

∴ The number in the standard form is 1 × 10^-8 g

Annual sales of sugar in sugar factory = 3 billion 720 million kilograms = 3720000 kg

∴ standard form of 3720000 kg = 372 × 10 × 10 × 10 × 10 kg

= 372 × 10⁴ kg

= 3.72 × 10⁴ × 10² kg

Using the law of exponent,

= 3.72 × 10⁶ kg

∴ The number in the standard form is 3.72 × 10⁶ kg

The number of red blood cells in blood = 5.5 million = 5500000 mm³

Blood in a body = 5 litres = 500000 mm³ (∵1 litre = 1,00,000 mm³)

∴ The total number of red cells in the body = 5500000 × 500000

= 55 × 5 × 10⁵ × 10⁵

Using the law of exponent,

= 275 × 10¹⁰

= 2.75 × 10² × 10¹⁰

= 2.75 × 10¹²

∴ The standard form is 2.75 × 10¹²

Given that, The mass of a proton =

Standard form =

= 1673 × 10^-27 g

= 1.673 × 10³ × 10^-27

Using the law of exponent,

= 1.673 × 10^{-27 + 3}

= 1.673 × 10^-24 g

∴ The standard form is 1.673 × 10^-24 g

Given that, The diameter of a helium atom = 0.000000022 cm

Standard form = 0.22 × 10^-7 cm

= 2.2 × 10^-1 × 10^-7

Using the law of exponent,

= 2.2 × 10^-1-7

= 2.2 × 10^-8 cm

∴ The standard form is 2.2 × 10^-8 cm

Given that, Mass of a molecule of hydrogen gas = 0.00000000000000000000334 tons

Standard form = 0.334 × 10^-20 tons

= 3.34 × 10^-1 × 10^-20

Using the law of exponent,

= 3.34 × 10^-1-20

= 3.34 × 10^-21

∴ The standard form is 3.34 × 10^-21 tons

Given that, cells in human body = 1 trillon

1 trillon = 1000000000000

Standard form = 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10

Using the law of exponent,

= 10¹²

∴ The standard form is 10¹²

Given that,56 km = 56 × 1000 m (∵ 1 km = 1000 m)

= 56000 m

Standard form = 56 × 10³

= 5.6 × 10¹ × 10³

Using the law of exponent,

= 5.6 × 10⁴ m

∴ The standard form is 5.6 × 10⁴ m

Given that,5 tons = 5 × 100 kg (∵ 1 ton = 100 kg)

= 5 × 100 × 1000 g (∵ 1 kg = 1000 g)

= 500000g

Standard form = 5 × 10 × 10 × 10 × 10 × 10

= 5 × 10⁵

∴ The standard form is 5 × 10⁵ g

Given that,2 years = 2 × 365 days (∵ 1 year = 365 days)

= 2 × 365 × 24 (∵ 1 day = 24 hours)

= 2 × 365 × 24 × 60 min (∵ 1 hr = 60 min)

= 2 × 365 × 24 × 60 × 60 s (∵ 1 min = 60 s)

= 63072000 s

Standard form = 63072 × 10 × 10 × 10

= 63072 × 10³

= 6.3072 × 10⁴ × 10³

Using the law of exponent,

= 6.3072 × 10⁷ s

∴ The standard form is 6.3072 × 10⁷ s

Given that, 5 hectares = 5 × 10000 m² (∵1 hectare = 10000 m²)

= 5 × 10000 × 100 × 100 cm²

Standard form = 5 × 10000 × 100 × 100

= 5 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10

Using the law of exponent,

= 5 × 10⁸ cm²

∴ The standard form is 5 × 10⁸ cm²

Given that,

Using the law of exponent,

On comparing two sides,

-3 = 2x-1

2x = -2

x = -1

∴ The value of x is -1.

Let as assume the number is x.

Let is divided by x to get the quotient

So,

⇒

Using the law of exponent,

Using the law of exponent,

∴The number is

Given that,

Using the law of exponent,

a^m ÷ aⁿ = a^m-n

⇒ 6^{n + 2} = 6³

On comparing both sides

n + 2 = 3

n = 1

∴ The value of n is 1

Given that,

Using the law of exponent,

⇒2ⁿ × 2⁶ × 2³ = 2¹⁸

Using the law of exponent,

⇒ 2^{n + 9} = 2¹⁸

On comparing both sides

n + 9 = 18

n = 9

∴ The value of n is 9

Given that,

Using the law of exponent,

⇒ = 5³ × 5³ × 5^-2 × x^-3 × x^-6

⇒ = 5⁴ × x³

= 5 × 5 × 5 × 5 × x³

= 625 x³

∴ The answer is 625 x³

⇒

⇒

⇒ 4×100

⇒ 400

⇒

⇒ 5^{m + 3-2-(-5)} = 5¹²

⇒ 5^{m + 3-2 + 5} = 5¹²

⇒ 5^{m + 6} = 5¹²

On comparing both sides,

m + 6 = 12

⇒ m = 12-6

⇒ m = 6

Weight of new born bear = 4 kg

Weight increases by the power of 2 in 5 years.

Weight of bear in 5 years = (4)²

= 16 kg

a. Cell of a bacteria in 30 mins = 2 (double in every 30 minutes)

So, cell of bacteria in 1 hour = 2² = 4

Cell of bacteria in 12 hours = = 2²⁴

b. Cell of bacteria in 24 hours = = 2⁴⁸

Distance between Planet A and Earth = 9.35 × 10⁶ km = 0.935 × 10⁷ km

Distance between Planet B and Earth = 6.27 × 10⁷ km

Clearly, planet A is nearer to Earth than planet B.

The cells of a bacteria double itself every hour = 1 + 1 = 2 = 2¹

Total number of cell in 8 hours = = 2⁸

a.

b.Distance covered in n hops = 1-

c. No, because for reaching 1, has to be equal to 0 which is not possible.

A. For numbers 2,3,7 and 8, pattern is of 4 digits.

For numbers 4 and 9, pattern is of 2 digits.

For numbers 1,5,6 and 10, pattern is of 1 digit.

B. 1. Pattern in number 4 is of 2 digits, 4 and 6.

On dividing power of 4, that is 12, by 2 and checking the remainder , we can predict the one’s digit in 4¹²

⇒

⇒ 4⁰ (on diving 12 by 2 we get remainder 0)

one’s digit is = 6 (second number out of 4 and 6)

2. Pattern in number 9 is of 2 digits, 9 and 1.

On dividing power of 9, that is 20, by 2 and checking the remainder , we can predict the one’s digit in 9²⁰

⇒

⇒ 9¹⁰ (on dividing 20 by 2 we get remainder 0)

one’s digit is = 1 (second number out of 9 and 1)

3. Pattern in number 3 is of 4 digits, 3,9,7 and 1.

On dividing power of 3, that is 17, by 4 and checking the remainder , we can predict the one’s digit in 3¹⁷

⇒

⇒ 3¹ (on dividing 17 by 4 we get remainder 1)

one’s digit is = 3 (first number out of 3,9,7 and 1)

4. Pattern in number 5 is of 1 digit, 5.

one’s digit is = 5

5. Pattern in number 10 is of 1 digit, 0.

one’s digit is = 0

C. 1. Pattern in number 1 is of 1 digit, 1.

one’s digit is = 1

2. Pattern in number 2 is of 4 digits, 2,4,8 and 6.

On dividing power of 12, that is 10, by 4 and checking the remainder , we can predict the one’s digit in 12¹⁰

⇒

⇒ 12² (on dividing 10 by 4 we get remainder 2)

one’s digit is = 4 (second number out of 2,4,8 and 6)

3. Pattern in number 7 is of 4 digits,7,9,3 and 1.

On dividing power of 17, that is 21, by 4 and checking the remainder , we can predict the one’s digit in 17²¹

⇒

⇒ 17¹ (on dividing 21 by 4 we get remainder 1)

one’s digit is = 7 (first number out of 7,9,3 and 1)

4. Pattern in number 9 is of 2 digits,9 and 1.

On dividing power of 29, that is 10, by 2 and checking the remainder , we can predict the one’s digit in 29¹⁰

⇒

⇒ 29⁰ (on dividing 10 by 2 we get remainder 0)

one’s digit is = 1 (last number out of 9 and 1)

(A)

B. Planets and the moon by mass, from least to greatest are:

(i) Pluto-1.27 × 10²²

(ii) Moon-7.35 × 10²²

(iii) Mercury-3.3 × 10²³

(iv) Venus-4.87 × 10²⁴

(v) Earth-5.97 × 10²⁴

(vi) Uranus-8.68 × 10²⁵

(vii) Neptune-1.02 × 10²⁶

(viii) Saturn-5.68 × 10²⁶

(ix) Jupiter-1.9 × 10²⁷

(x) Mars-6.42 × 10²⁹

(xi) Sun-1.99 × 10³⁰

C. Venus(4.87 × 10²⁴) has about the same mass as Earth(5.97 × 10²⁴).

(A)

(B) Planets from closest to the sun to farthest from the sun are:

(i) Mercury-5.7910⁷

(ii) Venus-1.08210⁸

(iii) Earth-1.496 × 10⁸

(iv) Jupiter-7.7810⁸

(v) Saturn-1.42710⁹

(vi) Uranus-2.8710⁹

(vii) Neptune-4.49710⁹

(viii) Pluto-5.910⁹

We know that as the negative exponent increases, the number becomes small. Smaller the negative exponent, larger the number.

A. Let us consider the exponents.

Lead and Silver have the same power.

Considering their decimals,

Lead is heavier than Silver as 3.44 is greater than 1.79.

∴ The heaviest element is Lead.

B. Let us consider the exponents.

Silver has 10^-25 whereas Titanium has 10^-26.

Titanium has a larger negative exponent.

∴ Titanium is lighter than Silver.

C. Let us consider the exponents of all the elements.

Hydrogen has the largest negative exponent, so it is the lightest.

Next, Titanium and Lithium have the same 10^-26.

Considering their decimals, Lithium is lighter, then comes Titanium.

Lastly, Lead and Silver have the same power 10^-25.

Considering their decimals,

Lead is heavier than Silver as 3.44 is greater than 1.79.

∴ The five elements in order of lightest to heaviest is

Hydrogen < Lithium < Titanium < Silver < Lead.

Given, Distance of Uranus from the Sun = 2, 896, 819, 200, 000 metres

⇒ 2, 896, 819, 200, 000 = 2.8968192 × 10¹²

∴ In standard form, planet Uranus is approximately 2.8968192 × 10¹² metres away from the sun.

Given 1 inch = 0.02543 metres

⇒ 0.02543 = =

We know that by laws of exponents, .

⇒ = 2.543 × 10^-2 metres

∴ In standard form, 1 inch = 2.543 × 10^-2 metres.

Given, volume of Earth is approximately 7.67 × 10^-7 times the volume of Sun.

We know by laws of exponents, a^-n =

Usual form:

⇒ 7.67 × 10^-7 = = = 0.000000767

∴ The volume of Earth is approximately 0.000000767 times the volume of Sun.

Given mass of electron = 9.1093826 × 10^-31 kilograms

We know that 1 kilogram = 1000 grams = 10³ grams.

⇒ Mass of electron = 9.1093826 × 10^-31 × 10³

We know that by properties of exponents, a^m × aⁿ = a^{m + n}.

⇒ 9.1093826 × 10^-31 × 10³ = 9.1093826 × 10^{-31 + 3}

= 9.1093826 × 10^-28

∴ Mass of electron in grams = 9.1093826 × 10^-28 grams

The world population at the end of 20^th century = 6.1 × 10⁹ people

Usual form of 6.1 × 10⁹ = 6.1 × 1, 000, 000, 000

= 6, 100, 000, 000

6, 100, 000, 000 can be read as six billion one hundred million.

A. Table showing number of ancestors in each of the 12 generations:

Graph showing number of ancestors in each of the 12 generations:

B. Let the number of ancestors be ‘a’.

From the table, we know that

⇒ a = 2⁰, 2¹, 2², 2³, 2⁴ … 2¹⁰, 2¹¹, 2¹².

∴ a = 2ⁿ where n is the generation

is the equation for number of ancestors in a given generation n.

Given litres of water that flow through the river in a day = 230 billion = 230, 000, 000, 000 = 230 × 10⁹ litres

We know that 1 week = 7 days and 1 year = 365 days.

Litres of water that flow through that river in a week = 230 × 10⁹ × 7 = 1610 × 10⁹

In standard notation, 1610 × 10⁹ = 1.61 × 10³ × 10⁹

We know that by properties of exponents, a^m × aⁿ = a^{m + n}.

⇒ 1.61 × 10³ × 10⁹ = 1.61 × 10^{3 + 9} = 1.61 × 10¹²

∴ Litres of water that flow through that river in a week = 1.61 × 10¹² litres

Litres of water that flow through that river in a year = 230 × 10⁹ × 365 = 83950 × 10⁹

In standard notation, 83950 × 10⁹ = 8.395 × 10⁴ × 10⁹

We know that by properties of exponents, a^m × aⁿ = a^{m + n}.

⇒ 8.395 × 10⁴ × 10⁹ = 8.395 × 10^{4 + 9} = 8.395 × 10¹

∴ Litres of water that flow through that river in a year = 8.395 × 10¹³ litres

Given, 300 grams of a substance decrease to 300 × 2^-3 after 3 half-lives.

⇒ Evaluating 300 × 2^-3

We know by laws of exponents, a^-n =

⇒ 300 × 2^-3 = = = = 37.5 grams

∴ 3.75 grams of the substance are left.

Given that the fraction of radioactive substance that remains after t half-lives can be found by the expression 3^-t.

A. Here, t = 7

∴ Fraction of substance that remains after 7 half-lives = 3^-7

We know by laws of exponents, a^-n =

⇒ 3^-7 = =

∴ of radioactive substance remains after 7 half-lives.

B. Fraction of substance that remains after t half-lives =

But fraction of radioactive substance that remains after t half-lives = 3^-t

∴ 3^-t =

243 can also be written as 3⁵.

⇒

We know that by laws of exponents, .

⇒ 3^-t = 3^-5

As bases are equal, we equate the powers.

⇒ -t = -5

∴ t = 5

∴ After 5 half-lives the fraction will be of the original.

Given 1 Fermi = 10^-15 m

Radius of proton = 1.3 Fermis

Radius of proton in metres = 1.3 × 10^-15

Moving 1 decimal to the right,

⇒ 1.3 × 10^-15 = 13 × 10¹ × 10^-15

We know that by properties of exponents, a^m × aⁿ = a^{m + n}.

⇒ 13 × 10¹ × 10^-15 = 13 × 10^1-15

= 13 × 10^-14

∴ The radius of a proton in standard form is 13 × 10^-14 m.

The length of the paper clip is 0.05 m.

There are 2 decimal places after the point.

⇒ 0.05 =

100 can be written as 10².

⇒

We know that by laws of exponents, .

⇒

∴ The length of the paper clip in standard form is 5 × 10^-2 m.

A.

⇒ 4 can be written as 2².

⇒

=

We know that by properties of exponents,

⇒

∴

B. 4^n-1

We know that by properties of exponents,

⇒ 4^n-1 =

=

∴ 4^n-1 =

C. 25 (5^n-2)

⇒ 25 can also be written as 5².

⇒ 25 (5^n-2) = 5² (5^n-2)

We know that by properties of exponents, a^m × aⁿ = a^{m + n}.

⇒ 5² (5^n-2) = 5^{2 + (n-2)}

= 5^{2 + n-2}

= 5ⁿ

∴ 25 (5^n-2) = 5ⁿ

First blank:

144 × 2^-3

We know by laws of exponents, a^-n =

⇒ 144 × 2^-3 =

=

= 18

Second blank:

18 × 12^-1

We know by laws of exponents, a^-n =

⇒ 18 × 12^-1 =

=

Third blank:

We know by laws of exponents, a^-n =

⇒

=

=

∴ 144 × 2^-3 = 18; 18 × 12^-1 = ; =

Given, there are 86, 400 seconds in a day.

1 day = 86400 seconds

∴ 1 second = of a day

⇒

Counting the number of places after the decimal point,

There are 9 places after the decimal point.

⇒ 0.000011574 =

⇒ 1000000000 can be written as 10⁹.

∴

We know that by law of exponents,

∴

= 1.1574 × 10^4-9

= 1.1574 × 10^-5 of a day

∴ A second is 1.1574 × 10^-5 of a day.

Let W₁ be the work done by (x2³)machine.

Let W₂ be the work done by (x3³) machine.

Work done by both the machines is W_t = W₁xW₂

W_t = 2³x3³

W_t = 2x2x2x3x3x3 = 8x27 = 216

If a single repeater has to be used, then the machine is of the form (x A^m) where A and m are both natural numbers.

216 = 6x6x6 = 6³

A = 6, m = 3.

Therefore, Shikha should use a (x6³) single repeater machine.

Let W₁ be the work done by firstmachine.

Let W₂ be the work done by second machine.

Work done by both the machines is W_t = W₁xW₂

W_t = 25² = 625

625 = 5x5x5x5 = 5²x5²

W₁xW₂ = 5²x5²

W₁ = W₂ = 5²

Therefore, Neha can use a hook-up of two (x5²) machines.

A. From the figure, it is evident that the input is 5cm and the output is 5cm. = 1. Therefore, it is a (x1) repeater.

B. From the figure, it is evident that the input is 3cm and the output is 15cm. = 5. Therefore, it is a (x5) repeater.

C. From the figure, it is evident that the input is 1.25cm and the repeater is a (x4). = 4, gives Output = 1.25x4 = 5cm.

D. From the figure, it is evident that the output is 36cm and the repeater is a (x4) and a (x3) = (x12). = 12, gives Input = = 3cm.

(a)

(x100)

Prime factorization of 100 = 2x2x5x5 = 2² x 5²

The hook-up is (x2²) and (x5²)

(b)

(x99)

Prime factorization of 99 = 3x3x11 = 3² x 11

The hook-up is (x3²) and (x11)

(c)

(x37)

37 itself is a prime number. A hook-up machine by definition must contain more than one machines. Since the factors of 37 are 1 and 37 and we are forbidden to use (x1) machines, no hook-up is possible.

(d)

(x1111)

Prime factorization of 1111 = 11x101

The hook-up is (x11) and (x101)

Let W₁ be the work done by firstmachine.

Let W₂ be the work done by second machine.

Work done by both the machines is W_t = W₁xW₂

W_t = 81 = 9x9

9x9 = 3x3x3x3

W₁xW₂ = 3²x3²

W₁ = W₂ = 3²

Therefore, the two repeater machines are (x3²) and (x3²).

Factorization of = =

Therefore, a (x) machine can do the same work as a (x ) machine.

By definition, a hook-up of machines consists of a combination of two or more machines.

(i) A repeater of 5x5 = 5²

(x5²) repeater.

(ii) A repeater of 5x5x5 = 5³

(x5³) repeater.

(iii) A repeater of 5x5x5x5 = 5⁴

(x5⁴) repeater.

Using the formula = Stretch Magnitude, we get

Using the formula = Repeater Magnitude

A chess board consist of 8 by 8 squares.

It is evident from the problem that the arrangement is in the powers of 2.

The first square has 2⁰ rupee coin.

The second square has 2¹ rupee coin.

The third square has 2² rupee coin, and so on.

Total sum S = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + … + 2⁶³ (64 terms)

This is in a series called a geometric progression, whose sum can be calculated using the formula S = where a is the first term, i.e 2⁰ = 1, n is the number of terms and r is the common ratio which is 2.

S = = Rs. 3,68,93,48,81,47,41,91,03,230.

Diameter Ratio

Therefore, The diameter of the sun is 1097.5 times that of earth.

Mass of Mars = 6.42 × 10²⁹ kg = 0.642 × 10³⁰ kg

Mass of Sun = 1.99 × 10³⁰ kg

Total Mass = (0.642 × 10³⁰ + 1.99 × 10³⁰) kg

= 2.6632x10³⁰ kg

From the problem, SE = 3.84 × 10¹¹ m

ME = 1.496 × 10⁸ m

SM = x m

SM = SE-ME

3.84 × 10¹¹ m - 1.496 × 10⁸ m

3.84 × 10¹¹ m – 0.001496 × 10¹¹ m

3.84 × 10¹¹ m – 0.001496 × 10¹¹ m

SM = 3.838 × 10¹¹ m

Speed = or

Time Taken =

=

= 2.7x10⁸ seconds.

We know that Dividend = (Quotient x Divisor) + Remainder.

Here, Dividend is (–15)^–1

Quotient = (–5)^–1, remainder assumed to be 0.

Therefore, (–15)^–1 = (–5)^–1 x Divisor.

Divisor = = =

Therefore, (–15)^–1 should be divided by (3)^-1 to get a quotient of (–5)^–1.

Let p be the number to be multiplied with (–8)^–3

(–8)^–3 x p = (–6)^–3

p =

Therefore, (–8)^–3 should be multiplied with for the product to be equal to (–6)^–3.

We know that, when the bases of the terms are same and those terms are divided, then their exponents are subtracted.

So,

As the bases are same so the exponents can be equated,

⇒ -5 + 7 = -x

⇒ x = -2

We know that, when the bases of the terms are same and those terms are multiplied, then their exponents are added.

So,

As the bases are same so the exponents can be equated,

⇒ 2x + 9 = x + 2

⇒ x = 2 – 9 = -7

∴ The value of x is -7

From the above given equation, we take 2^x common, we get

3( 2^x ) = 192

⇒ 2^x = 64

⇒ 2^x = 2⁶

∵ 2⁶ is equal to 64

As the bases are same so the exponents can be equated,

⇒ x = 6

∴ value of x = 6

As we know that, any number with power zero equals to 1, i.e a⁰=1,

where a = any number

Therefore, expressing the given equation, in exponent format,

As the bases are same so the exponents can be equated,

⇒ x – 7 = 0

⇒ x = 7

The above given equation can also be written as,

2^3x = ((2)³)^{2x + 1} ∵ 2³ = 8

As we know by the laws of exponent, that,

(a^m)ⁿ = a^mn

⇒ 2^3x = 2^{3(2x + 1)}

As the bases are same so the exponents can be equated,

⇒ 3x = 3(2x + 1)

⇒ x = 2x + 1

⇒ x = -1

∴ Value of x =-1

Given: 5^x+ 5^x–1 = 750

We know that, when the bases of the terms are same and those terms are multiplied, then their exponents are added.

⇒ 5^x + 5^x ×5^-1 = 750

By taking 5^x from above equation, we get,

5^x (1 + 5^-1) = 750

As when the bases of the terms are same and those terms are divided, then their exponents are subtracted.

⇒ 5^{x – 1} = 5³

As the bases are same so the exponents can be equated,

⇒ x – 1 = 3

⇒ x = 4

By substituting the values of a and b in given equation, we get,

⇒ (-1)² + (2)^-1

As we know by the property of negative exponents,

∴ the value of given expression is .

By substituting the values of a and b in given equation, we get,

⇒ (-1)² - (2)^-1

As we know by the property of negative exponents,

∴ the value of given expression is

By substituting the values of a and b in given equation, we get,

⇒ (-1)² × (2)²

⇒ 1 × 4 = 4

∴ the value of given expression is 4

By substituting the values of a and b in given equation, we get,

⇒ (-1)² ÷ (2)^-1

As we know by the property of negative exponents,

∴ the value of given expression is 2

On making the factors of 1296, we get,

1296 = 2⁴ × 3⁴

Similarly, we make factors of 14641, we get,

⇒ 14641 = 11⁴

As we know that, 125 = 5³ also 343 = 7³,

On making the factors of 400 we get,

400 = 2⁴ × 5²

Similarly, we make factors of 3969, we get,

⇒ 3969 = 3⁴ × 7²

On making the factors of 625 we get,

625 = 5⁴

Similarly, we make factors of 10000, we get,

⇒ 10000 = 2⁴ × 5⁴

Given equation can be written as:-

The given equation can be written as:

On multiply the given equation we get,

We know that, when the bases of the terms are same and those terms are multiplied, then their exponents are added.

∴ The value of given expression is 0

As we know that 9 = 3² and 27 = 3³

Therefore, the given expression can also be written as:-

We know that, when the bases of the terms are same and those terms are multiplied, then their exponents are added.

We also know that, when the bases of the terms are same and those terms are divided, then their exponents are subtracted.

⇒ 3⁷ × t²

∴ The value of the given expression is 3⁷ × t²

As (a^m)ⁿ = a^mn , therefore, the given expression can also be written as:-

We know that, when the bases of the terms are same and those terms are divided, then their exponents are subtracted.

⇒ (3)^{10 – 4} × 5⁰ × t^{12 – 6}

⇒ 3⁶ × t⁶ ∵ a⁰ = 1

∴ The value of the given expression is 3⁶ × t⁶