12 Exponents And Powers
12.1 Introduction
12.2 Powers With Negative Exponents
12.3 Laws Of Exponents
12.4 Use Of Exponents To Express Small Numbers In Standard Form
12.4.1 Comparing Very Large And Very Small Numbers
Numbers with negative exponents obey the following laws of exponents.
(a) am × an = am+n
(b) am an = am–n
(c) (am)n = amn
(d) am × bm = (ab)m
(e) a0 = 1
(f) a^m/b^m= (a/b)m
Example 1:
Find the value of
(i) 2–3
(ii) (1/3)^-2
Example 2: Simplify
(i) (– 4)5 × (– 4)–10 (ii) 25 ⎟ 2– 6
Solution: (i) (– 4)5 × (– 4)–10 = (– 4) (5 – 10) = (– 4)–5 = 5
(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
−=
(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
Solution:
(i) (25 ⎟ 28)5 × 2– 5 = (25 – 8)5 × 2– 5 = (2– 3)5 × 2– 5 = 2– 15 – 5 = 2–20 = 20 2
(ii) (– 4)– 3 × (5)– 3 × (–5)–3 = [(– 4) × 5 × (–5)]– 3 = [100]– 3 = 3 100
[using the law am × bm = (ab)m, a–m=1m
= (–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
Therefore, m + 6 = 7 or m = 7 – 6 = 1
Example 6: Find the value of 232
EXERCISE 12.1
1. Evaluate.
(i) 3–2
(ii) (– 4)– 2
(iii) 1/2– 5
2. Simplify and express the result in power notation with positive exponent.
(i) (– 4)5 ⎟ (– 4)8
(ii) (1/2³)²
(iii) (-3)
3. Find the value of.
4
(iv) (3– 7 ⎟ 3– 10) × 3– 5
(v) 2– 3 × (–7)– 3
(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)
8 5
⋅(ii) (5–1 × 2–1) × 6–1
5. Find the value of m for which 5m ⎟ 5– 3 = 55. − 6. Evaluate (i) 1314
(ii)
7. Simplify.
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.
1. Write the following numbers in standard form.
(i) 0.000000564 (ii) 0.0000021 (iii) 21600000 (iv) 15240000 2. Write all
the facts given in the standard form.
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 7.54 7.54
Again we need to convert numbers in standard form into
(ii) 7.54 × 10– 4 = 4
= = 0.000754
10 10000 3 3
a numbers with the same exponents.
(iii) 3 × 10– 5 = 5
= = 0.00003
10 100000
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 1/1000000 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?
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