Tiêu đề 4. indices.pdf ✅

Chapter Kinematics 12 Figure 1.1 Chapter Indices 4 REmEmBER Before beginning this chapter, you should be able to: • Recognise terms like constant, variable, etc. • Have thorough knowledge of mathematical operators KEy IDEaS After completing this chapter, you should be able to: • Understand the terms like exponentiation, index, etc. • Explain the laws of indices • Solve exponential equations • Use exponents on larger numbers M04 IIT Foundation Series Maths 7 9019 05.indd 1 1/30/2018 7:56:42 PM
4.2 Chapter 4 INTRODUCTION The sum of n instances of a number x is n times x, denoted as nx, i.e., multiplication is repeated addition. Multiplying by n means adding n instances of a number. For example, n + n = 2 × n n + n + n = 3 × n n + n + n + ... x times = x × n The product of n instances of a number x is denoted as xn. This is an exponential expression, read as x to the power of n or x raised to n and the operation (of multiplying n instances of x) is called exponentiation or involution. Exponentiation is a repeated multiplication. Raising x to the power of n (where n is a natural number) is the product of n instances of x. For example, n × n = n2 n × n × n = n3 n × n × n × ... x times = nx In the expression nx, n is called the base and x is the index (plural: indices) or exponent. The entire expression is called the xth power of n. For example, 6 × 6 × 6 × 6 × 6 × 6 × 6 can be written as 67. Here, 6 is the base and 7 is the index (or exponent). We shall look at the rules/properties pertaining to these exponential expressions in this chapter. Initially, we shall consider only the positive integral values of the indices. We shall also see what meaning we can assign to exponential expressions in which the index is 0 or a negative integer or a rational number. In higher classes, we shall consider all real values of the index. As for the base itself, it can be any real number. We shall first verify the following laws for positive integral values of the index. Laws of Indices 1. am × an = am + n (Product of powers) For example, (a) 35 × 37 = 35 + 7 = 312 (b) 3 4 3 4 3 4 3 4 3 4 6 3 6 1 3 10       ×       ×       =       =       + + (c) 52 × 54 × 57 × 510 × 53 = 52 + 4 + 7 + 10 + 3 = 526 (d) 3 3 3 3 4 7 4 7 11 ( ) × ( ) = ( ) = ( ) + Note a a a a a m m 1 1 2 3 m mn n m m2 3 m m × × × = + + + + ... ... 2. am ÷ an = am – n, a ≠ 0 (Quotient of powers) For example, (a) 96 ÷ 92 = 96 – 2 = 94 (b) 4 5 4 5 4 5 4 5 12 7 12 7 5       ÷       =       =       − Note We now consider what meaning we can assign to a0. M04 IIT Foundation Series Maths 7 9019 05.indd 2 1/30/2018 7:56:43 PM
Indices 4.3 If we want these laws to be true for all values of m and n, i.e., even for n = m, from above (2), we get a a a m m m m = − or 1 = a0 We see that if we define a0 as 1, then this law will be true even for n = m. Therefore, we define a0 as 1, provided a ≠ 0. When a = 0, an – n = an/an = 0/0, which is not defined. ∴ 00 is not defined. Note We shall now consider, what meaning we can assign to an, when n is a negative integer. We have am × an = am + n. Consider an × a−n = an + (−n) (if we want the law to be true) = a0 = 1. ∴ an × a–n = 1 ⇒ a–n = 1 an and 1 a−n = an. If we define a−n as 1 an , then this law is true even for negative value of n. For example, 4−3 = 1 43 , 3−1 = 1 3 and a−m = 1 am (provided a ≠ 0). Note a b a b b a       = = −1 1 3. (am) n = am × n (power of a power) For example, (a) 7 7 7 3 4 3 4 12 ( ) = = × (b) 6 7 6 7 6 7 3 5 3 5 15               =       =       × Note [(am)n]p = amnp and so on 4. (ab)n = an × bn (power of a product) For example, (a) (21)4 = (3 × 7)4 = 34 × 74 (b) (210)9 = (2 × 3 × 5 × 7)9 = 29 × 39 × 59 × 79 In problems, we may often want to write an × bn as (ab)n. For example, (a) 32 × 243 = 25 × 35 = (2 × 3)5 = 65 (b) 64 125 27 1331 × = 4 5 3 11 3 3 3 3 × = 4 5 3 11 3 3       ×       = 4 5 3 11 3 ×       = 12 55 3       Note (a b c d ... z)n = an bn c n dn ... zn 5. a b n n (Power of a quotient) For example, (a) 3 4 3 4 5 5 5       = (b) 12 7 4 3 7 4 3 7 4 3 7 6 6 6 6 6 6 6       =  ×      = ( ) × = × In problems, we may want to write down a b n n as a b n       . M04 IIT Foundation Series Maths 7 9019 05.indd 3 1/30/2018 7:56:50 PM
4.4 Chapter 4 For example, (a) 5 7 5 7 4 4 4 =       (b) 6 7 5 3 6 7 5 3 18 35 5 5 5 5 / / / / ( ) ( ) =       =       6. a b = b a -n n             For example, (a) 7 8 8 7 4 4       =       − (b) 1 3 3 1 3 1 1       =       = − Note 1 1 1 a a a       =       = − Example 4.1 Which is greater between 51014 and 9676? Solution Let us take the GCD of 1014 and 676, which is 338. 51014 can be written as (53)338, i.e., (125)338. Similarly, 9676 can be written as (92)338 = (81)338. Clearly, 51014 > 9676 Example 4.2 What should be multiplied to 6–2 so that the product may be equal to 216? Solution Let the required number be x. ∴ x × 6–2 = 216 ⇒ x × 1 62 = 63 ( ∴ a–m = 1/am) ⇒ x = 63 × 62 ⇒ x = 63 + 2 ( ∴ am × an = am + n) ⇒ x = 65 ∴ The required number is 65. Exponential Equation 1. If am = an, then m = n, if a ≠ 0, a ≠ 1, and ≠ −1. For example, (a) If 4n = 42, then n = 2. (b) If 3q = 81, i.e., 3q = 34, then q = 4. M04 IIT Foundation Series Maths 7 9019 05.indd 4 1/30/2018 7:56:52 PM