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# ch07 - Chapter 7 The Logarithmic Function and Logarithms...

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Review of Prerequisite Skills 4. The increase in population is given by f ( x ) = 2400(1.06) x f (20) = 2400(1.06) 20 = ˙ 7697. The population in 20 years is about 7700. 5. The function representing the increase in bacteria population is f ( t ) = 2000(2 4 t ). Determine t when f ( t ) = 512 000: 512 000 = 2000(2 4 t ) 256 = 2 4 t 4 t = 8 t = 32. The bacteria population will be 512 000 in 32 years. 6. a. The function is A ( t ) = 5 1 2 16 t 20 . When t = 200, A ( t ) = 5 1 2 1 2 6 0 2 0 0 = 4.59. They will have 4.59 g in 200 years. b. Determine t when A ( t ) = 4. 4 = 5 1 2 16 t 20 1 2 16 t 20 = 0.8 16 t 20 log 0.5 = log 0.8 t = 1620 l l o o g g 0 0 . . 8 5 = 521.52 Exercise 7.1 4. 5. 6. a. Let log 2 8 = x . Then 2 x = 8, by definition. But 2 3 = 8. x = 3 So, log 2 8 = 3. c. Let log 3 81 = x . 3 x = 81 x = 4 So, log 3 81 = 4. e. Let log 2 1 8 = x . 2 x = 1 8 2 x = 2 –3 x = –3 So, log 2 1 8 = –3. – 5 5 1 1 5 y = x y y = log x 1 5 x y 5 – 5 – 5 5 1 y = x x y = log 5 x y = 5 x Chapter 7: The Logarithmic Function and Logarithms 163 Chapter 7 • The Logarithmic Function and Logarithms

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d. log 3 4 27 = log 3 (27) 1 4 = log 3 (3 3 ) 1 4 = log 3 3 3 4 = 3 4 e. log 3 (9 5 9) = log 3 (3 2 9 1 5 ) = log 3 (3 2 3 2 6 ) = log 3 3 1 5 2 = 1 5 2 by definition of logarithms f. log 2 16 1 3 = log 2 (2 4 ) 1 3 = log 2 2 4 3 = 4 3 8. b. log 4 x = 2 x = 4 2 x = 16 d. log 4 6 1 4 = x 4 x = 6 1 4 4 x = 4 –3 x = –3 f. log 1 4 x = –2 x = 1 4 –2 = 4 2 = 16 164 Chapter 7: The Logarithmic Function and Logarithms g. Let log 5 5 = x . 5 x = 5 5 x = 5 1 2 x = 1 2 So, log 5 5 = 1 2 . i. Let log 2 4 32 = x. 2 x = 4 32 2 x = (32) 1 4 2 x = (2 5 ) 1 4 2 x = 2 5 4 x = 5 4 So, log 2 4 32 = 5 4 . 7. a. log 6 36 – log 5 25 = 2 – 2 = 0 b. log 9 1 3 + log 3 1 9 = log 9 (3 –1 ) + log 3 (9 –2 ) = log 9 [ (9) 1 2 ] –1 + log 3 [(3 2 )] –2 = log 9 9 1 2 + log 3 (3 –4 ) = – 1 2 + (–4) = – 4 1 2 or 9 2 or – 4.5 c. log 6 36 – log 25 5 = log 6 (36) 1 2 – log 25 (25 1 2 ) = log 6 6 – log 25 (25 1 2 ) = 1 – 1 2 = 1 2
9. To find the value of a logarithm, you can use the button on a calculator for those which have base 10. The log can be rounded to the number of decimals appropriate to the problem. If, however, the number can be expressed as a power of the base of the logarithm, then the exponent of the power is the logarithm. Since 16 can be written as 2 4 , log 2 16 = log 2 2 4 , = 4. By definition log b b x = x . If the number cannot be so expressed, we can use the calculator to find log a b by finding l l o o g g 1 1 0 0 b a i.e., log 2 16 = l l o o g g 10 1 1 0 2 6 = 4 by calculator. 10. x y ±4 81 8 4 1 ±3 27 2 1 7 ±2 9 1 9 ±1 3 1 3 0 2 11. Integer values for y exist where x is a power of 10 with an integer exponent. If y > –20, the smallest number is 10 –19 . If x 1000, the largest number is 1000 = 10 3 . There are integers from –19 to 3 that satisfy the condition, so there are 3 – (–19) + 1 = 23 integer values of y possible. – 4 4 x y 2 catenary LOG Chapter 7: The Logarithmic Function and Logarithms 165 Section 7.2 Investigation: 2. log b 1 = 0 3. log b b = 1 4. log b b x = x 5. b log b x = x Exercise 7.2 6. a. y = log 3 x b. y = log 3 (9 x ) = log 3 9 + log 3 x = 2 + log 3 x c. y = log 3 (27 x ) = log 3 27 + log 3 x = 3 + log 3 x d. y = log 3 3 x = log 3 x – 1 7. a. y = log 3 x b. y = log 3 x 2 = 2log 3 x c. y = log 3 x 3 = 3log 3 x 3 3 – 3 y x 6 ( ) log log log x log x 3

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e. log 2 224 – log 2 7 = log 2 22 7 4 = log 2 32 = 5 f. log 3 36 + log 3 3 4 = log 3 36 3 4 = log 3 27 = 3 9. a. log 3 3 + log 5
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