Chapter 12

Exponents and Powers

12.1 Introduction

Do you know?

Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have already learnt in earlier class how to write such large numbers more conveniently using exponents, as, 5.97 × 1024 kg.

We read 10²⁴ as 10 raised to the power 24.

We know 2⁵ = 2 × 2 × 2 × 2 × 2

and 2m = 2 × 2 × 2 × 2 × ... × 2 × 2 ... (m times)

Let us now find what is 2– 2 is equal to?

12.2 Powers with Negative Exponents

You know that, 102 = 10 × 10 = 100

101 = 10 =

100 = 1 =

10– 1 = ?

Continuing the above pattern we get, 10– 1 =

Similarly 10– 2 =

10– 3 =

What is 10– 10 equal to?

Now consider the following.

33 = 3 × 3 × 3 = 27

32 = 3 × 3 = 9 =

31 = 3 =

3° = 1 =

So looking at the above pattern, we say

3– 1 = 1 ÷ 3 =

3– 2 = = =

3– 3 = = × =

You can now find the value of 2– 2 in a similar manner.

We have, 10– 2 = or 102 =

10– 3 = or 103 =

3– 2 = or 32 = etc.

In general, we can say that for any non-zero integer a, a– m = , where m is a positive integer. a–m is the multiplicative inverse of am.

We learnt how to write numbers like 1425 in expanded form using exponents as

1 × 103 + 4 × 102 + 2 × 101 + 5 × 10°.

Let us see how to express 1425.36 in expanded form in a similar way.

We have 1425.36 = 1 × 1000 + 4 × 100 + 2 × 10 + 5 × 1 +

= 1 × 103 + 4 × 102 + 2 × 10 + 5 × 1 + 3 × 10– 1 + 6 × 10– 2

10– 1 = , 10– 2 =

TRY THESE

We have learnt that for any non-zero integer a, am × an = am + n, where m and n are natural numbers. Does this law also hold if the exponents are negative? Let us explore.

(i) We know that 2 – 3 = and 2 – 2 =

  for any non-zero integer a.

Therefore, = 2 – 5

–5 is the sum of two exponents – 3 and – 2

(ii) Take (–3)– 4 × (–3)–3

(– 4) + (–3) = – 7

(iii) Now consider 5–2 × 54

5–2 × 54 = = 5(2)

(iv) Now consider (–5)– 4 × (–5)2

(–5)– 4 × (–5)2 =

= = (–5)– (2)

In general, we can say that for any non-zero integer a,

am × an = am + n, where m and n are integers.

TRY THESE

On the same lines you can verify the following laws of exponents, where a and b are non zero integers and m, n are any integers.

(iv) (v) a0 = 1

Let us solve some examples using the above Laws of Exponents.

(i) 2–3 (ii)

Solution:

(i) (ii)

Example 2: Simplify

(i) (– 4)5 × (– 4)–10 (ii) 25 ÷ 2– 6

Solution:

(i) (– 4)5 × (– 4)–10 = (– 4) (5 – 10) = (– 4)–5 = (am × an = am + n, )

(ii) 25 ÷ 2– 6 = 25 – (– 6) = 211 (am ÷ an = am – n)

Example 3: Express 4– 3 as a power with the base 2.

Solution: We have, 4 = 2 × 2 = 22

Therefore, (4)– 3 = (2 × 2)– 3 = (22)– 3 = 22 × (– 3) = 2– 6 [(am)n = amn]

Example 4: Simplify and write the answer in the exponential form.

(i) (25 ÷ 28)5 × 2– 5 (ii) (– 4)– 3 × (5)– 3 × (–5)– 3

(iii) (iv)

Solution:

(i) (25 ÷ 28)5 × 2– 5 = (25 – 8)5 × 2– 5 = (2– 3)5 × 2– 5 = 2– 15 – 5 = 2–20 =

(ii) (– 4)– 3 × (5)– 3 × (–5)–3 = [(– 4) × 5 × (–5)]– 3 = [100]– 3 =

[using the law am × bm = (ab)m, a–m=]

(iii)

(iv) = = (–1)4 × 34 ×

= (–1)4 × 54 = 54 [(–1)4 = 1]

Example 5: Find m so that (–3)m + 1 × (–3)5 = (–3)7

Solution: (–3)m + 1 × (–3)5 = (–3)7

(–3)m + 1+ 5 = (–3)7

(–3)m + 6 = (–3)7

On both the sides powers have the same base different from 1 and – 1, so their exponents must be equal.

Therefore, m + 6 = 7

or m = 7 – 6 = 1

Example 6: Find the value of .

Solution:

Example 7: Simplify (i)

(ii)

Solution:

(i) =

=

(ii) =

=

Exercise 12.1

1. Evaluate.

(i) 3–2 (ii) (– 4)– 2 (iii)

2. Simplify and express the result in power notation with positive exponent.

(i) (– 4)5 ÷ (– 4)8 (ii)

(iii) (iv) (3– 7 ÷ 3– 10) × 3– 5 (v) 2– 3 × (–7)– 3

3. Find the value of.

(i) (3° + 4– 1) × 22 (ii) (2– 1 × 4– 1) ÷ 2– 2 (iii)

(iv) (3– 1 + 4– 1 + 5– 1)0 (v)

4. Evaluate (i) (ii) (5–1 × 2–1) × 6–1

5. Find the value of m for which 5m ÷ 5– 3 = 55.

6. Evaluate (i) (ii)

7. Simplify.

(i) (ii)

12.4 Use of Exponents to Express Small Numbers in Standard Form

Observe the following facts.

1. The distance from the Earth to the Sun is 149,600,000,000 m.

2. The speed of light is 300,000,000 m/sec.

3. Thickness of Class VII Mathematics book is 20 mm.

4. The average diameter of a Red Blood Cell is 0.000007 mm.

5. The thickness of human hair is in the range of 0.005 cm to 0.01 cm.

6. The distance of moon from the Earth is 384, 467, 000 m (approx).

7. The size of a plant cell is 0.00001275 m.

8. Average radius of the Sun is 695000 km.

9. Mass of propellant in a space shuttle solid rocket booster is 503600 kg.

10. Thickness of a piece of paper is 0.0016 cm.

11. Diameter of a wire on a computer chip is 0.000003 m.

12. The height of Mount Everest is 8848 m.

Observe that there are few numbers which we can read like 2 cm, 8848 m,
6,95,000 km. There are some large numbers like 150,000,000,000 m and some very small numbers like 0.000007 m.

Identify very large and very small numbers from the above facts and write them in the adjacent table:

We have learnt how to express very large numbers in standard form in the previous class.

For example: 150,000,000,000 = 1.5 × 1011

Now, let us try to express 0.000007 m in standard form.

0.000007 = = = 7 × 10– 6

0.000007 m = 7 × 10– 6 m

Similarly, consider the thickness of a piece of paper which is 0.0016 cm.

0.0016 =

= 1.6 × 10– 3

12.4.1 Comparing very large and very small numbers

The diameter of the Sun is 1.4 × 109 m and the diameter of the Earth is 1.2756 × 107 m. Suppose you want to compare the diameter of the Earth, with the diameter of the Sun.

Diameter of the Sun = 1.4 × 109 m

Diameter of the earth = 1.2756 × 107 m

Therefore = = which is approximately 100

So, the diameter of the Sun is about 100 times the diameter of the earth.

Let us compare the size of a Red Blood cell which is 0.000007 m to that of a plant cell which is 0.00001275 m.

Size of Red Blood cell = 0.000007 m = 7 × 10– 6 m

Size of plant cell = 0.00001275 = 1.275 × 10– 5 m

Therefore, = = (approx.)

So a red blood cell is half of plant cell in size.

Mass of earth is 5.97 × 1024 kg and mass of moon is 7.35 × 1022 kg. What is the
total mass?

Total mass = 5.97 × 1024 kg + 7.35 × 1022 kg.

= 5.97 × 100 × 1022 + 7.35 × 1022

= 597 × 1022 + 7.35 × 1022

= (597 + 7.35) × 1022

= 604.35 × 1022 kg.

When we have to add numbers in standard form, we convert them into numbers with the same exponents.

The distance between Sun and Earth is 1.496 × 1011m and the distance between Earth and Moon is 3.84 × 108m.

During solar eclipse moon comes in between Earth and Sun.

At that time what is the distance between Moon and Sun.

Distance between Sun and Earth = 1.496 × 1011m

Distance between Earth and Moon = 3.84 × 108m

= 1.496 × 1000 × 108 – 3.84 × 108

= (1496 – 3.84) × 108 m = 1492.16 × 108 m

Example 8: Express the following numbers in standard form.

(i) 0.000035 (ii) 4050000

Solution: (i) 0.000035 = 3.5 × 10– 5 (ii) 4050000 = 4.05 × 106

Example 9: Express the following numbers in usual form.

(i) 3.52 × 105 (ii) 7.54 × 10– 4 (iii) 3 × 10– 5

Solution:

(i) 3.52 × 105 = 3.52 × 100000 = 352000

(ii) 7.54 × 10– 4 = = 0.000754

(iii) 3 × 10– 5 = = 0.00003

Exercise 12.2

1. Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942

(iii) 6020000000000000 (iv) 0.00000000837

(v) 31860000000

2. Express the following numbers in usual form.

(i) 3.02 × 10– 6 (ii) 4.5 × 104 (iii) 3 × 10– 8

(iv) 1.0001 × 109 (v) 5.8 × 1012 (vi) 3.61492 × 106

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

(i) 1 micron is equal to m.

(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

(iii) Size of a bacteria is 0.0000005 m

(iv) Size of a plant cell is 0.00001275 m

(v) Thickness of a thick paper is 0.07 mm

4. In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack.

What have we Discussed?