NCERT Solutions Class 7 Math Chapter 11 Exponents and Powers Example

Example 1. Express 256 as a power 2.

Solution – We have 256 = 2x2x2x2x2x2x2x2.
So we can say that 256 = 2⁸

Example 2. Which one is greater 2³ or 3²?

Solution – We have,

2³=2x2x2=8 and
3²=3×3=9.
Since 9>8, so, 3² is greater than 2³

Example 3. Which one is greater 8² or 2⁸

Solution – 8² = 8 x8=64
2⁸ = 2x2x2x2x2x2x2x2 = 256

Clearly, 2⁸ > 8²

Example 4. Expand a³b², a²b³, b²a³, b³a². Are they all same?

Solution – a3b² = a³x b²
= (a x a x a) x (b x b)
= a x a x a x b x b
be xaxa)x(bxb)

a²b³ = a² x b³
= a x a x b x b x b

b² a² = b² x a³
= b x b x a x a x a

b³ a² = b³ x a²
= b x b x b x a x a

Note that in the case of terms a³ b² and a² b³ the powers of a and b are different. Thus a³ b² and a b³ are different.

On the other hand, a³ b² and b² a³ are the same, since the powers of a and b in these two terms are the same. The order of factors does not matter.

Thus, a³ b²=a³ xb² = b² x a³ = b² a³. Similarly, a b³ and b³ a² are the same.

Example 5. Express the following numbers as a product of powers of prime factors:

(i) 72
(ii) 432
(iii) 1000
(iv) 16000

Solution –

(i) 72 = 2 x 36 = 2 x 2 x 18
= 2 × 2× 2 × 9
= 2 x 2 x 2 x 3 x 3 = 2³ x 3²

Thus, 72 = 2³ x 3² (required prime factor product form)

(ii) 432 = 2 x 216 = 2 × 2 × 108 = 2 x 2 x 2 x 54
= 2 x 2 x 2 x 2 x 27 = 2 x 2 x 2 x 2 × 3 × 9
= 2 x 2 x 2 x 2 x 3 x 3 x 3

or 432 = 2⁴ x 3³ (required form)

1000 = 2 × 500 = 2 x 2 x 250 = 2 x 2 x 2 x 125
= 2 x 2 x 2 x 5 x 25 = 2 x 2 x 2 x 5 x 5 x 5
or 1000 = 2³ x 5³

Atul wants to solve this example in another way:

1000 = 10 x 100 = 10 x 10 x 10
= (2 × 5) x (2 × 5) x (2×5) (Since10 = 2×5)
= 2 x 5 x 2 x 5 x 2 x 5 = 2 x 2 x 2 x 5 x 5 x 5
or 1000 = 2³ x 5³

Is Atul’s method correct?

(iv) 16,000 16 x 1000 = (2 x 2 x 2 x 2) x 1000 = 2⁴ x 10³ (as 16 = 2 x 2 x 2 x 2)
= (2 × 2 × 2 × 2) × (2 × 2 × 2 × 5 x 5 x 5) = 2⁴ × 2³ × 5³

(Since 1000 = 2 x 2 x 2 x 5 x 5 x 5)
= (2 × 2 × 2 × 2× 2 x 2 x 2) x (5 x 5 x 5)

or, 16,000 = 2⁷ x 5³

Example 6. Work out (1)⁵, (-1)³, (-1)⁴, (-10)³, (-5)⁴

Solution –
(i) We have (1)⁵ = 1 x 1 x 1 x 1 x 1 = 1
In fact, you will realise that 1 raised to any power is 1.

(ii) (-1)³ = (-1) x (-1) x (-1) = 1 × (-1) = -1

(iii) (-1)⁴ = (-1) × (-1) × (−1) × (−1) = 1 ×1=1

You may check that (-1) raised to any odd power is (-1), and (-1) raised to any even power is (+1).

(iv) (-10)³ = (-10) (-10) x (-10) = 100 x (-10) = – 1000

(v) (-5)⁴ = (-5) x (-5) × (-5) × (-5)= 25 x 25 = 625

Example 7. Can you tell which one is greater (5²) × 3 or (5²)³?

Solution – (5²) × 3 means 5² is multiplied by 3 i.e., 5 × 5 × 3 =75

but (5²)³ means 5² is multiplied by itself three times i.e.,

5² x 5² x 5²=5⁶ = 15,625
Therefore (5²)³> (5²) x 3

Example 8. Express the following terms in the exponential form:

(i) (2×3)⁵
(ii) (2a)⁴
(iii) (-4m)³

Solution –

(i) (2×3)⁵ = (2×3) × (2×3) × (2×3)x(2×3)x(2×3)
= = (2×2×2×2×2) × (3 × 3× 3 × 3 × 3)
= 2⁵ x 3⁵

(ii) (2a)⁴ += 2a x 2a x 2 x 2a
= (2×2×2×2) x (axaxaxa)
= 2⁴ x a⁴

(iii) (-4m)³= (-4xm)³
= = (-4xm) x (-4xm) x (-4xm)
= (-4)x(4) × (4) × (m × m × m) = (-4)³ × (m)³

Example 9. Expand:

(i) (3/5)⁴
(ii) (-4/7)⁵

Solution –

(i) (3/5)⁴ = 3⁴/5⁴ = 3x3x3x3/5x5x5x5

(ii) (-4/7)⁵ = (-4)⁵/7⁵ = (-4)× (-4)× (-4)× (-4) × (-4)/7×7×7×7×7

Numbers with exponent zero

Can you tell what 3⁵/3⁵ equals to?

3⁵/3⁵ = 3x3x3x3x3/3x3x3x3x3 = 1

by using laws of exponents

And

3⁵ ÷ 3⁵ = 3^5-5 = 3⁰
So 3⁰ = 1

Can you tell what 7° is equal to?

7³ ÷ 7³ = 7^3-3 = 7⁰

And 7³ ÷ 7³ = 7x7x7/7x7x7 = 1

Therefore 7⁰ = 1

Similarly a³ ÷a³ = a^3-3 = a⁰

And a³ ÷ a³ = a³/a³ = axaxa/axaxa = 1

Thus a⁰ = 1 (for any non-zero integer a)

So, we can say that any number (except 0) raised to the power (or exponent) () is 1.

Solution –

We have, 8x8x8x8 = 8⁴

But we know that 8 = 2 x 2 x 2 = 2³

Therefore 8⁴ = (2³)⁴ = 2³ x 2³ x 2³ x 2³

= 2^3×4 [You may also use (a^m)ⁿ =a^mn]

= 212

Example 11. Simplify and write the answer in the exponential form.

(i) (3⁷/3²) x 3⁵
(ii) 2³ x 2² x 5⁵
(iii) (6²x6⁴)÷ 6³
(iv) [(2²)³ x 3⁶] x 5⁶
(v) 8² ÷ 2³

Solution –

(i) (3⁷/3²) x 3⁵ = (3^7-2) x 3⁵
= 3⁵ x 3⁵ = 3⁵⁺⁵ = 3¹⁰

(ii) 2³ x 2² x 5⁵ = 2³⁺² x 5⁵
= 2⁵ x 5⁵ = (2 × 5)⁵ = 10⁵

(iii) (6²x6⁴) ÷ 6³ = 6²⁺⁴ ÷ 6³
= 6⁶/6³ = 6^6-3 = 6³

(iv) [(2²)³ x 3⁶] x 5⁶ = [2⁶ x 3⁶] x 5⁶
= (2×3)⁶ x 5⁶
= (2x3x5)⁶ – 30⁶

(v) 8 = 2x2x2 = 2³
Therefore 8² ÷ 2³ = (2³)² ÷ 2³
= 2⁶ ÷ 2³ = 2^6-3 = 2³

Example 12. Simplify:

(i) 12⁴ x 9³ x 4/6³ x 8² x 27
(ii) 2³ x a³ x 5a⁴
(iii) 2 x 3⁴ x 2⁵/9 x 4²

Solution –
(i) We have

12⁴ x 9³ x 4/6³ x 8² x 27 = (2²x3)⁴ x(3²)³ x 2²/(2×3)³ x (2³)² x 3³

= (2²)⁴ x (3)⁴ x 3^2×3 x 2²/2³ x 3³ x 2^2×3 x 3³ = 2⁸ x 2² x 3⁴ x 3⁶/2³ x 2⁶ x 3³ x 3³

= 2⁸⁺² x 3⁴⁺⁶/2³⁺⁶ x 3³⁺³ = 2¹⁰ x 3¹⁰/2⁹ x 3⁶

= 2^10-9 x 3^10-6 = 2¹ x 3⁴

= 2 x 81 = 162

(ii) 2³ × a³ × 5a⁴ = 2³ x a³ × 5 x a⁴
= 2³ x 5 x a³ x a⁴ = 8 x 5 x a³⁺⁴
= 40 a⁷

(iii) 2 x 3⁴ x 2⁵/9 x 4² = 2 x 3⁴ x 2⁵ \3²× (2²)² =2×2⁵×3⁴ \3²×2^2×2

=2¹⁺⁵ ×3⁴\2⁴×3⁴ = 2⁶×3⁴\2⁴×3² =^26-4×3^4-2

=2²×3²=4×9 =36

Note: In most of the examples that we have taken in this Chapter, the base of a power was taken an integer. But all the results of the chapter apply equally well to a base
which is a rational number.

(i) 5985.3
(ii) 65,950
(iii) 3,430,000
(iv) 70,040,000,000

Solution –

(i) 5985.35.9853 x 1000 = 5.9853 × 10³
(i) 65,950 = 6.595 × 10,000 = 6.595 x 10⁴
(iii) 3,430,000 = 3.43 x 1,000,000 = 3.43 x 10⁶
(iv) 70,040,000,000 = 7.004 x 10,000,000,000 = 7.004 × 10¹⁰

A point to remember is that one less than the digit count (number of digits) to the left of the decimal point in a given number is the exponent of 10 in the standard form. Thus, in 70,040,000,000 there is no decimal point shown; we assume it to be at the (right) end. From there, the count of the places (digits) to the left is 11. The exponent of 10 in the standard form is 11 – 1 = 10. In 5985.3 there are 4 digits to the left of the decimal point and hence the exponent of 10 in the standard form is 4-1 = 3.