Exponential and Logarithmic Functions

Transcription

1 Chapter 6 Eponential and Logarithmic Functions 6.3 Logarithmic Functions. 9 = 3 is equivalent to = log = 4 is equivalent to = log a =.6 is equivalent to = log a.6 4. a 3 =. is equivalent to 3 = log a. 5.. = M is equivalent to = log. M = N is equivalent to 3 = log. N 7. = 7. is equivalent to = log = 4.6 is equivalent to = log = π is equivalent to = log π 0. π = e is equivalent to π = log e. e = 8 is equivalent to = ln8. e. = M is equivalent to. = ln M 3. log 8 = 3 is equivalent to 3 = 8 4. log 3 ( 9) = is equivalent to 3 - = 9 5. log a 3=6 is equivalent to a 6 = 3 6. log b 4 = is equivalent to b = 4 7. log 3 = is equivalent to 3 = 8. log 6 = is equivalent to = 6 9. log M =.3 is equivalent to.3 = M 0. log 3 N =. is equivalent to 3. = N. log π = is equivalent to ( ) = π. log π = is equivalent to π = 3. ln4 = is equivalent to e = 4 4. ln = 4 is equivalent to e 4 = 5. log = 0 since 0 = 6. log 8 8 = since 8 = 8 7. log 5 5 = since 5 = 5 8. log 3 ( 9) = since 3 = 9 9. log 6 = 4 since ( ) 4 = 4 = log 9 = since ( 3) = 3 = log 0 0 = since 0 3 = 0 3. log 5 5 = since 5 3 = = 5 594

2 Section 6.3 Logarithmic Functions 33. log 4 = 4 since ( ) 4 = log 3 9 = 4 since ( 3) 4 = ln e = since e = e 36. ln e3 = 3 since e 3 = e The domain of f () = ln( 3) is: 3 > 0 > 3 > 3 { } 39. The domain of F() = log is: > 0 0 { } 4. The domain of h() = log + ( ) is: + > 0 ( ) > 0 { } 43. The domain of f () = ln + is: + > 0 + > 0 > > { } 38. The domain of g() = ln( ) is: > 0 > > { } 40. The domain of H() = log 5 3 is: 3 > 0 > 0 > 0 { } 4. The domain of G() = log ( ) is: > 0 ( +)( ) > 0 < or > < or > { } 44. The domain of g() = ln 5 is: 5 > 0 5 > 0 > 5 > 5 { } 45. The domain of g() = log requires that > 0. The epression is zero or undefined when = or = 0. f () = + Interval Test Number Positive/Negative << / Positive <<0 0.5 Negative 0<< Positive The domain is < or > 0 { } 595

3 Chapter 6 Eponential and Logarithmic Functions 46. The domain of h() = log 3 requires that > 0. The epression is zero or undefined when = 0 or =. f () = Interval Test Number Positive/Negative <<0 / Positive 0<< 0.5 Negative << Positive The domain is { < 0 or > }. 47. ln 5 3 = ln ln(0 / 3) 0.04 = ln( / 3) For f () = log a, find a so that f () = log a = or a = or a =. (The base a must be positive b definition.) 5. For f () = log a, find a so that f ( ) = log a ( ) = 4. a 4 = a = 4 a 4 4 = a = (The base a must be positive b definition.) 53. B 54. F 55. D 56. H 57. A 58. C 59. E 60. G 6. f () = ln( + 4) Using the graph of = ln, shift the graph 4 units to the left. Domain: ( 4, ) Vertical Asmptote: = 4 596

4 6. f () = ln( 3) Using the graph of = ln, shift the graph 3 units to the right. Domain: (3, ) Vertical Asmptote: = 3 Section 6.3 Logarithmic Functions 63. f () = ln( ) Using the graph of = ln, reflect the graph about the -ais. Domain: (,0) Vertical Asmptote: = f () = ln( ) Using the graph of = ln, reflect the graph about the -ais, and reflect about the -ais. Domain: (,0) Vertical Asmptote: = g() = ln( ) Using the graph of = ln, compress the graph horizontall b a factor of. Vertical Asmptote: = 0 597

5 Chapter 6 Eponential and Logarithmic Functions 66. h() = ln ( ) Using the graph of = ln, stretch the graph horizontall b a factor of. Vertical Asmptote: = f () = 3ln Using the graph of = ln, stretch the graph verticall b a factor of 3. Vertical Asmptote: = f () = ln Using the graph of = ln, stretch the graph verticall b a factor of, and reflect about the -ais. Vertical Asmptote: = g() = ln(3 ) = ln( ( 3) ) Using the graph of = ln, reflect the graph about the -ais, and shift 3 units to the right. Domain: (,3) Vertical Asmptote: = 3 598

6 Section 6.3 Logarithmic Functions 70. h() = ln(4 ) = ln( ( 4) ) Using the graph of = ln, reflect the graph about the -ais, and shift 4 units to the right. Domain: (,4) Vertical Asmptote: = 4 7. f () = ln( ) Using the graph of = ln, shift the graph unit to the right, and reflect about the -ais. Domain: (, ) Vertical Asmptote: = 7. f () = ln Using the graph of = ln, reflect the graph about the -ais, and shift units up. Vertical Asmptote: = log 3 = = 3 = log ( + ) = 3 + = 3 + = 8 = 7 = log 5 = 3 = 5 3 = log 3 (3 )= 3 = 3 3 = 9 3 = = log 4 = = 4 = (, base is positive) 78. log 8 ( ) = 3 3 = 8 = 599

7 Chapter 6 Eponential and Logarithmic Functions 79. ln e = 5 e = e 5 = 5 8. log 4 64 = 4 = 64 4 = 4 3 = ln e = 8 e = e 8 = 8 = 4 8. log 5 65 = 5 = 65 5 = 5 4 = log 3 43 = = = = 5 = 4 = 85. e 3 = 0 3 = ln0 = ln log 6 36 = = = = 5 = = e = 3 = ln 3 = ln e + 5 = = ln8 = 5 + ln8 = 5 + ln8 89. log 3 ( +) = + = 3 + = 9 = 8 9. log 8 = 3 8 = 3 8 = 8 8 = 8 = = ± 8 = ± 88. e + = 3 + = ln3 = + ln3 = + ln3 90. log 5 ( + + 4) = = = 5 + = 0 = ± = log 3 3 = 3 = 3 = 4()( ) () or = + 85 ±

8 Section 6.3 Logarithmic Functions 93. f ( ) = ( ) if < 0 ln ln if > 0 Domain: (, 0) ( 0, ) Range: (, ) -intercept: ( -, 0), (, 0) vertical asmptote: = f( ) = ln ln ( ) if ( ) if -<<0 Domain: (, 0) Range: [ 0, ) -intercept ( -, 0) vertical asmptote: = f( ) = ln if 0< < ln if Range: [ 0, ) -intercept: (, 0) vertical asmptote: = 0 60

9 Chapter 6 Eponential and Logarithmic Functions 96. f( ) = ln if 0 < < ln if Range: (,0] -intercept: (, 0) vertical asmptote: = P = 00e 0. n (a) 50 = 00e 0. n (b) 5 = 00e 0.n 0.5 = e 0. n 0.5 = e 0. n ln0.5= 0.n ln0.5= 0.n n = ln0.5 n = ln n 6.93 n panes of glass are needed. 4 panes of glass are needed. 98. ph = log 0 [ H + ] (a) ph = log 0 [ ] = ( 7) = 7 (b) 4. = log 0 [ H + ] 4. = log 0 H + [ ] H w = 50e d (a) 30 = 50e d (b) 5 = 50e d 0.6 = e d 0. = e d 00. A = A 0 e 0.35 n ln0.6= d d = ln d 7.7 Approimatel 8 das. [ ] = 0 4. = = ln0.= d d = ln d Approimatel 576 das. (a) 50 = 00e 0.35 n (b) 0 = 00e 0.35 n 0.5 = e 0.35 n 0. = e 0.35 n ln0.5= 0.35 n n = ln das 0.35 ln0.= 0.35 n n = ln das 0.35 Approimatel das. Approimatel 6.6 das. 60

10 Section 6.3 Logarithmic Functions 0. F(t) = e 0. t (a) 0.5 = e 0. t (b) 0.8 = e 0. t 0.5 = e 0. t 0.5 = e 0. t ln0.5= 0.t t = ln n 6.93 Approimatel 7 minutes. 0. = e 0. t 0. = e 0. t ln0.= 0.t t = ln0. 0. n 6.09 Approimatel 6 minutes. (c) It is impossible for the probabilit to reach 00% because e 0. t will never equal zero. 0. F(t) = e 0.5 t (a) 0.50 = e 0.5t (b) 0.80 = e 0.5 t 0.5 = e 0.5 t 0. = e 0.5 t 0.5 = e 0.5 t ln0.5= 0.5t t = ln minutes Approimatel 5 minutes. 0. = e 0.5 t ln0.= 0.5t t = ln minutes 0.5 Approimatel minutes. ( ) ( ) 03. D = 5e 0.4 h 04. N = P e 0.5 d = 5e 0.4 h 0.4 = e 0.4 h ln0.4= 0.4h h = ln h.9 hours 05. I = E R R e L t 0.5 ampere:.0 ampere: 0.5 = 0 e = e t 0 t 5 e t = t = ln t = ln t = seconds 450 = 000 e 0.5 d 0.45 = e 0.5 d 0.55 = e 0.5 d 0.55 = e 0.5 d ln0.55= 0.5d.0 = 0 0 t e = e t e t = t = ln0.667 t = ln0.667 t = seconds d = ln das 0.5 I seconds t 603

11 Chapter L(t) = A e k t Eponential and Logarithmic Functions ( ) k (5) ( ) (a) 0 = 00 e (b) (c) (d) 0. = e 5 k e 5 k = 0.9 5k = ln0.9 k = ln L(0) = 00 e 0.0(0) ( ) = 00( e 0. ) = ( ) = 00( e 0.35 ) = L(5) = 00 e 0.0(5) ( ) 0.9 = e 0.0 t e 0.0 t = = 00 e 0.0 t 0.0t = ln0. t = ln das ( ) 38 words ( ) 54 words 07. (a) R = 3e k 0 = 3e k ( 0.06 ) 0 3 = e k ( 0.06) ln 0 = k( 0.06) 3 ln 0 3 k = (b) R = 3e ( 0.07) ( 0.7) R = 3e % (c) 00 = 3e ( 0.07) 00 3 = e ( 0.07 ) ln 00 = ( 0.07) 3 ln 00 3 = 0.07 ( ) (d) (e) ( 5 = 3e 0.07 )( ) 5 = e ( 0.07 )( ) ( ) = ( 0.07) ( ) ( ) ln 5 = ln Answers will var. 08. Answers will var. 604

12 Section 6.3 Logarithmic Functions 09. New = Old( e R t ) Age Depreciation rate Age Depreciation rate = 36600e R ( ) = 3400e R ) = e R = R R = 3.8% = e R = R = R R = 8% Age Depreciation rate Age Depreciation rate = 8750e R 3) = 5400e R 4) = e 3 R = 3R = R 3 R = 9.3% Age Depreciation rate = 00e R ( 5 ) = e 5 R = 5R = R R = e 4 R = 4 R = R 4 R = 0.% 605