# 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 } 1 2 } 2 } 16 t 20 } . When t = 200, A ( t 1 } 1 2 } 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 } 1 2 } 2 } 16 t 20 } 1 } 1 2 } 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 ,by definition. But 2 3 . x =3 So, log 2 3 . c. Let log 3 81 = x . 3 x =81 x =4 So, log 3 81 = 4. e. Let log 2 1 } 1 8 } 2 = x . 2 x = } 1 8 } 2 x =2 –3 x =–3 So, log 2 1 } 1 8 } 2 = –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 Ï 4 27 w =log 3 (27) } 1 4 } 3 (3 3 ) } 1 4 } 3 3 } 3 4 } = } 3 4 } e. log 3 (9 3 Ï 5 9 w ) 3 (3 2 3 9 } 1 5 } ) 3 (3 2 3 3 } 2 6 } ) 3 3 } 1 5 2 } = } 1 5 2 } by definition of logarithms f. log 2 16 } 1 3 } 2 (2 4 ) } 1 3 } 2 2 } 4 3 } = } 4 3 } 8. b. log 4 x =2 x =4 2 x =16 d. log 4 1 } 6 1 4 } 2 = x 4 x = } 6 1 4 } 4 x –3 x =–3 f. log } 1 4 } x =–2 x = 1 } 1 4 } 2 –2 2 164 Chapter 7: The Logarithmic Function and Logarithms g. Let log 5 Ï 5 w = x . 5 x = Ï 5 w 5 x =5 } 1 2 } x = } 1 2 } So, log 5 Ï 5 w = } 1 2 } . i. Let log 2 Ï 4 32 w = x. 2 x = Ï 4 32 w 2 x =(32) } 1 4 } 2 x =(2 5 ) } 1 4 } 2 x } 5 4 } x = } 5 4 } So, log 2 Ï 4 32 w = } 5 4 } . 7. a. log 6 36 – log 5 25 – 2 =0 b. log 9 1 } 1 3 } 2 +log 3 1 } 1 9 } 2 9 (3 –1 )+log 3 (9 –2 ) 9 [ (9) } 1 2 } ] –1 3 [(3 2 )] –2 9 9 } 1 2 } 3 (3 –4 ) =– } 1 2 } +(–4) =–4 } 1 2 } or } 9 2 } or – 4.5 c. log 6 Ï 36 w – log 25 5 6 (36) } 1 2 } – log 25 (25 } 1 2 } ) 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. xy ±4 81 } 8 4 1 } ±3 27 } 2 1 7 } ±2 9 } 1 9 } ±1 3 } 1 3 } 02 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.

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## This note was uploaded on 02/10/2010 for the course MATHS MCV-01 taught by Professor Mr.m during the Spring '10 term at Seneca.

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

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