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STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 CHAPTER 2: STANDARD FORM, INDEX AND LOGARITHMS 1.0 INTRODUCTION Understanding index and logarithms is important because it is used widely in research, science, finance and engineering. For example, it can be used to show how substances are formed, multiply and decay in the natural world. Index Expressions Figure 1.1 shows our planet Earth orbiting the Sun. The distance between the Earth and the Sun is about 93 million kilometers. You can write this number as 93,000,000 km. Figure 1.1: Distance between Earth and Sun This number is obviously long to write and hard to read. You can also write this number as 9.3 x 107km. Numbers written this way is called indices. Usually only numbers that are too small and too big are stated in index form. Referring to the number 9.3 x 10 7, the number 10 is called the base and the number 7 is called the index. 1 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 1012 = 10 x 10 …………10 . 12 times base index In general, an means a multiplied by itself n times. An index expression is in simple form if there is: No repeating base No negative index For example x2y5 x 4 , a2 b –6 and p3 2 can be simplified as x6y5, 2 a2 p9 6 and b Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Activity 1.0 TRY THESE QUESTIONS! 1. Rewrite these numbers in index form: i. ii. a. 2000000 c. 0.00082 b. 138000000000 d. 0.000000015 2. The rate of reproduction of a particular insect is about 1430000000 a month. Write this in index. 3. First, express 400 kilometers in centimeters. Next, state this in index. 3 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 FEEDBACK for Activity 1.0 1. a. 2 x 106 a. c. 8.2 x 10-4 2. 1.43 x 109 3. b. 1.38 x 1011 d. 1.5 x 10-8 40 000 000 cm, 4 x 107 cm 4 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS 1.1 BA101/CHAPTER2 INDEX RULE There will be times when you need to add, subtract, multiply or divide two or more index numbers. The rules provided in Table 1.1 are most useful. Study it carefully. Given that a, b, m and n are real numbers. Rule Statements 1. Multiplication am x an = a m + n 2. Division am an = a m - n 3. Power i. ( a m ) n = a mn ii. (ab)n = an b n n an a iii. n ; b 0 b b 4. Negative Index i. a –n = a m b n ; a 0 and b 0 b n a m ii. 5. Zero Index 6. Fraction Index 1 ;a0 an a0 = 1 ; a0 m a n n am Table 1.1: Rules of Index Operations 5 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Example 1.1 Simplify the following expressions. a. 34 x 35 c. (48 )3 e. b. 85 82 d. ( 25 ) -3 70 f. 27 2 3 SOLUTION: a. b. c. d. e. f. Expressions 34 x 35 85 82 ( 48 )3 ( 25 ) -3 a0 = 1 70 27 Index Rule Used am x an = a m + n am an = a m - n i. ( a m ) n = a mn i. ( a m ) n = a mn 1 ii. a –n = n ; a 0 a 2 3 a m n a n Solutions 34 + 5 = 3 9 85 – 2 = 8 3 424 ( 2 5 ) -3 = 2 -15 1 2 –15 = 15 2 ; a0 1 2 3 27 3 27 2 9 m Example 1.2 A sum of RM10,000 is saved in a bank at 8% interest compounded monthly. The total sum J 0.08 after t years is given as, J = 100001 12 a. 6 6 months 12t . What is the total sum after b. 5 years Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 SOLUTION a. 6 months = 0.5 year. Therefore t = 0.5. 0.08 J = 100001 12 b. 6 t = 5. 0.08 J = 100001 12 7 = 10406.73 12( 5 ) =14898.46 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Activity 1.1 TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION……..! 1 Simplify a. c. 2. a5 x a 6 z 2 y5 x y –2 Simplify a. m 12 m 3 c. z 7 x z 6 z 5 3 Simplify a. ( x 3 ) 5 c. ( 2x2 y 3 z )5 e. 3 ( ab 2 ) –4 8 b. d. b. 3 x 3 8 x 3-4 4n x 16 2n x 32 –2n x 8 -n 2 8 24 d. 25 n 5 –2n x 125 2n b. ( 3x 4) 2 d. ( 10 3 ) 4 f. ( 2m 2 ) ( 4n )3 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Feedback for Activity 1.1 4. 5. 6. a. a11 c. z2 y 3 b. 35 a. m 9 c. z 8 b. 24 a. x 15 c. 32 x 10 y 15 z 5 b. 9x 8 3 e. 4 8 a b 9 d. 2 –3n d. 5 10n d. 1012 m2 f. 32n 3 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS 1.2 BA101/CHAPTER2 LOGARITHMIC EXPRESSIONS In this section, you will learn about the relationship between indices and logarithms. Let’s now consider the reproduction rate of amoeba, a single cell living organism that reproduces by replicating itself. If one new amoeba needs one day to replicate itself into 2 amoebas, there will be 4 amoebas after 2 days, 8 amoebas after 3 days, and so on. Time in days(x) Number of amoeba (y) 0 1 1 2 2 4 3 8 4 16 5 32 6 64 7 128 The index equation y = 2x can be used to represent the rate of reproduction of this amoeba. Conversely, this equation can also be written as x = log2 y , a logarithmic equation. log2 y is read as log of y to the base 2 log is a short form of logarithm 10 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS Index Form BA101/CHAPTER2 Logarithmic Form index y = ax x = log a y base Note: Observe that the base is still the same after changing index form to logarithmic form. Generally, if a is a positive number and y = a x, then x is equals to the logarithm of y to the base a. If y = a x , then x = log a y If x = log a y , then y = a x Equation 1.1 Example 1.3 Rewrite the following numbers in logarithmic form, (base given). a. 100 base10 b. 64 base 4 c. 64 base 2 d. 125 base 5 e. 81 base 3 Sample solutions a. If 100 = 10 2, then 2 = log 10 100 b. If 64 = 4 3, then 3 = log 4 64 c. If 64 = 2 6, then 6 = ? 11 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 d. If 125 = 5 3, then ? = log 5 125 e. If 81 = 3 4, then 4 = log ? 81 Example 1.4 Determine these values: a. log 7 49 b. log 2 0.5 c. log 9 3 d. log 10 0.001 Solution: a. Assuming then therefore b. Assuming then therefore c. Assuming then therefore d. Assuming then therefore 12 log 7 49 = x 49 = 7x 72 = 7 x x = 2 log 2 0.5 = x 0.5 = 2 x 0.5 = ½ = 2 – 1 = 2 x x = -1 log 9 3 = x 3 = 9x 3 = ( 3 2 )x 31 = 32x 1 = 2x ½ = x or x = 0.5 log 10 0.001 = x 0.001 = 10 x 1 1 3 10 3 10 x 1000 10 x = -3 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Activity 1.2 TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION……..! 1. Given that 32 = 25, determine the value of log 2 32 2. Given that 1/8 = 2 –3 , determine the value of log 2 1/8 3. Given that 8 = 64 , determine the value of log 64 8 4. Given that 0.001 = 10 –3 , determine the value of log 10 0.001 5. Calculate the value of a. log 3 81 c. log 8 4 b. log 7 343 d. log 27 9 6. Convert the following into logarithmic form a. 32 = 25 b. 50 = 10 1.699 c. a = x2 d. x-3 = 0.3 7. Find the value of x given that 13 a. log 3 1 = x b. log 7 49 = x c. log 10 0.001 = x d. log 5 25 = x Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Feedback for Activity 1.2 1. 5 2. –3 3. ½ 4. –3 5. a. x = 4 c. x = 2/3 6. a. log 2 32 = 5 c. log x a = 2 7. a. x = 0 c. x = -3 14 b. 3 d. 2/3 b. log 10 50 = 1.699 d. log x 0.3 = -3 b. x = 2 d. x = 2 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS 1.3 BA101/CHAPTER2 LOGARITHM RULE As in index, you will need to perform basic algebra operations on numbers in logarithm. Table 1.2 shows the logarithm rule. Assume M and N are positive real numbers 1. 2. 3. 4. 5. log a MN = log a M + log a N log a M/ N = log a M – log a N log a (M) c = c log a M log a a = 1 log a a0 = 0 log a a = 0 6. log N M = log a M log a N ( to convert base N to base a) 7. a log a N = N 8. log101 = log a (for base 10 only) Table 1.2: Rules of Logarithm Operations Example 1.5 Write the following expressions as addition or subtraction of logarithms. a. log a x 2 y 3 c. log 1 100b 2 b. log a 3 b 3/2 c. log ( ab 3 ) c Solution: a. log a x 2 y 3 = log a x2 + log a y 3 = 2 log a x + 3 log a y RULE 1 and 3 b. log a 3 b 3/2 = log a 3 + log b 3/2 3 = 3 log a + log b 2 15 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS c. log BA101/CHAPTER2 1 = log 1 – ( log 100 + log b2 ) 2 100b = 0 – log 100 – 2 log b = - ( log 100 + 2 log b ) = -2 (1 + log b ) 1/ 2 ab 3 ab 3 = ½ ( log ab3 – log c ) log d. log ( )= c c 3 = ½ log ab – ½ log c = ½ log a + ½ log b3 – ½ log c 3 = ½ log a + log b – ½ log c 2 Example 1.6 Rewrite the expressions below as a single logarithm a. log 2 + log 3 b. log 4 + 2 log 3 – log 6 c. 2 log x + 3 log y – log z d. ½ log a + log a2 b – 2 log ab Solution: a. log 2 + log 3 = log ( 2 x 3 ) = log 6 b. log 4 + 2 log 3 – log 6 = log 4 + log 32 – log 6 = log ( 4 x 9 ) – log 6 36 = log log 6 6 c. 2 log x + 3 log y – log z = log x2 + log y 3 – log z x2 y3 = log z d. ½ log a + log a2 b – 2 log ab = log a 1/2 + log a 2 b – log ( ab )2 log 16 a a1 / 2 a 2 b = log 2 b (ab) Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Example 1.7 Given log 2 = 0.3010 and log 3 = 0.4771 , find the value of: a. log 8 b. log 18 c. log 0.6 Solution: a. log 8 = log 2 3 = 3 log 2 = 3 ( 0.3010 ) = 0. 903 b. log 18 = log (2 x 9 ) = log 2 + log 9 = log 2 + log 3 2 = log 2 + 2 log 3 = 0.3010 + 2 ( 0.4771) = 1.2552 c. log 0.6 = log ( 6 ) 10 = log 6 – log 10 = log ( 2 x 3 ) – 1 = log 2 + log 3 – 1 = 0.3010 + 0.4771 – 1 = - 0 . 2219 17 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Activity 1.4 TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION……..! 1. Express the following as a combination of log x, log y or log z a. log x 3 y 2 b. log c. log xy x3 y2 d. log x2 y 3 z 2. Express the following as single logarithm a. log5 14 – log5 21 + log5 6 1 c. 4 log 3 2 log 3 25 2 3 log 7 9 2 log 7 6 2 2 8 d. 2 log 5 log 5 3 9 b. 3. Determine the values of a. log4 9 – log436 b. 2 log 2 5 – log2 100 + 3 log2 4 4. Given that log a 3 = 0.477 and log a 5 = 0.699, find the value of a. log a 15 c. 18 log a 45 log a 9a b. log a 35 d. loga 0.6 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Feedback for Activity 1.4 1. a. 3 log x + 2 log y b. 3 log x – 2 log y c. ½ log x + ½ log y d. 2 log x + log y – 3 log z 2 2. a. log5 4 b. log7 ¾ c. log3 80 d. log5 ½ 3. a. –1 b. 4 4. a. 1.176 b. 0.8265 c. 0.846 d. –0.222 19 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS 1.4 BA101/CHAPTER2 INDEX AND LOGARITHMIC EQUATIONS Congratulations! You have already understood and know how to use Index Rule and Logarithm Rule to solve simple problems involving Index and Logarithm expressions. Now we move on to solving simple equations involving Index and Logarithm. The following rule is very important. Assuming that x and y are real numbers If If ax = ay , then x = y log a x = log a y , then x = y Example 1.8 Solve the following equations a. 7x = 12 b. 3 5x – 8 = 9 x + 2 Solution: a. 7x = 12 Log both sides, x log 7 = log 12 log 7 0.845 x 0.783 log 12 1.079 b. 3 5x – 8 = 9 x + 2 3 5x – 8 = (3 2 ) x + 2 Similar base 3 5x – 8 = 3 2 x + 4 ( am )n = amn 5x – 8 = 2x + 4 If ax = ay , then x = y 5x – 2x = 4 + 8 3x = 12 x=4 20 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Example 1.9 Solve the following logarithmic equations a. log 2 ( 7 + x ) = 3 b. log 50 + log x = 2 + log ( x – 1 ) c. log 2 x = log x 16 Solution: a. log 2 (7 + x ) = 3 (7+x )= 23 7+x=8 x=1 b. log 50 + log x = 2 + log ( x – 1 ) log 50x = log 100 + log (x – 1 ) log 50x = log 100(x – 1 ) 50x = 100x – 100 100 = 50 x x=2 c. log 2 x = log x 16 log 2 16 log 2 x log 2 x Convert to index form log = log base 10 log a + log b = log ab 2 = log 10 100 Convert to similar base (log 2 x )2 = log 2 16 (log 2 x )2 = 4 (log 2 x ) = 2 log 2 x = 2 or log 2 x = -2 x = 2 2 or x = 2 – 2 x = 4 or ¼ 21 Convert to index form Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 Activity 1.5 TEST YOURSELF BEFORE YOU MOVE ON TO THE NEXT SECTION……..! 1. Solve these equations 2 a. 32x = 8 b. 2x – 3 = 4x + 1 c. 3 4x = 27 x – 3 d. 5 x 2562 x 2 Solve these equations a. log 6 x + log 6 ( x + 5 ) = 2 b. 5 log x6 - log x 96 = 4 c. 2 log 3 + log 2x = log ( 3x + 1 ) d. log 25 – log x + log ( x + 1 ) = 3 22 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 FEEDBACK for Activity 1.5 1 a. 32x = 8 (25)x = 23 5x = 3 3 x = 5 b. x = -5 c. x = -9 d. x = –6 or 2 2. 23 a. x = 4 or -9 b. x = 3 1 c. x = 15 1 d. x = 39 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 SELF ASSESSMENT 1 Another round of congratulations to you for making it so far. You are very close to mastering this unit. Attempt all questions in this section and check your solutions with the answers provided in SOLUTIONS TO SELF ASSESSMENT given after this. If you face any problems you cannot solve, please discuss with your lecturer. 1. Solve the following equations: a. 10 x = 0.00001 c. log 2 4x = 8 log x 2 2. b. 3 x 9 x – 1 = 27 2x –1 d. 5 log x 6 – log x 96 = 4 Simplify the expressions below: a. 3 log 2 5 10 1 2 log 2 log 2 3 9 30 b. log 3 5 log 5 2 log 2 3 3. Given that log 10 5 = p, express the following in term of p. a. log 10 250 c. log 5 10 4. b. log 10 0.5 d. log 5 1000 Solve the following equations: a. 3 log 2 + log ( 4x – 1 ) = log ( 7 – 8x ) b. 2 log 15 + log ( 5 – x ) – log 4x = 2 c. log 5 x 2 = 1 + log 5 ( 25 – 4x ) d. log 2 y 2 = 3 + log 2 ( y + 6 ) 24 Prepared By : Azmanira Muhamed STANDARD FORM, INDEX AND LOGARITHMS BA101/CHAPTER2 SOLUTIONS TO SELF ASSESSMENT 1. a. x = -5 1 c. x = or 4 16 b. x = 1/3 d. x = 3] e. 3 log 2 + log ( 4x – 1 ) – log ( 7 – 8x ) = 0 2. a. -3 b. –0.954 3. a. 2p + 1 b. p – 1 c. 1/p d. 3/p 4. a. x = 3 5 9 5 c. x = 5 or -25 d. y = 12 or -4 b. x = 25 Prepared By : Azmanira Muhamed