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Chapter 4

Chapter 4. Section 4.6 Exponential and Logarithmic Equations. Chapter 4. Section 4.4 Exponential and Logarithmic Equations. P(t). t P( t). 15. 14. 13. 12. 11. 10. 9. 8. 7. 6. 5. 4. 3. 2. t. 1950. 1960. 1970. 1980. 1990. 2000. 2010. 2020. 2030. 2040. 2050.

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Chapter 4

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  1. Chapter 4 Section 4.6 Exponential and Logarithmic Equations

  2. Chapter 4 Section 4.4 Exponential and Logarithmic Equations

  3. P(t) t P( t) 15 14 13 12 11 10 9 8 7 6 5 4 3 2 t 1950 1960 1970 1980 1990 2000 2010 2020 2030 2040 2050 Solving Equations: Examples 1. World population (x109) P3  P(t) = 3(1.018)t–1960 1950 2.5098 1960 3.0000 1970 3.5859 1980 4.2862 1990 5.1233 2000 6.1239  2010 7.3199 P2 2020 8.7495 2030 10.458  P1  2040 12.500 2050 14.942 Section 4.6 v5.0.1

  4. = 72,282,704 people/yr 6.12396 – 2.50982 3.61414 = = P = 176,369,423 people/yr 2000 – 1950 50 t P(2050) – P(2000) 2050 – 2000 Solving Equations: Examples 1. World population P(t) (billions) in year t is given by P(t) = 3(1.018)t–1960 What was the population in 1950 ? In 2000 ? P(1950) = 3(1.018)1950–1960 P(2000) = 3(1.018)2000–1960 What is the growth factor ? What is the averageannualincrease from 1950 to 2000? From 2000 to 2050 ? = 3(1.018)–10 = 2.50982 billion = 3(1.018)40 = 6.12396 billion 1.018 Section 4.6 v5.0.1

  5. P(2000) – P(1950) 14.9424 – 6.1239 6.1239 – 2.5098 P(2050) – P(2000) 143.999 % 143.999 % (100) (100) (100) (100) = = = = P(2000) 2.5098 P(1950) 6.1239 Solving Equations: Examples 1. World population (continued) What is theper cent change in population from 1950 to 2000 ? What is the per cent change in population from 2000 to 2050 ? Exponential Fact: On equal intervals exponentialgrowth functions always grow at a rate proportional to the initial value on the interval NOTE: The same is true for exponentialdecay functions Section 4.6 v5.0.1

  6. P(t) t P( t) 15 14 13 12 11 10 9 8 7 billion/yr 3.6141 8.8184 P P = = 6 t t 50 50 P P 5 (100) (100) 143.99 % 143.99 % = = P1 P2 4 3 2 t 1950 1960 1970 1980 1990 2000 2010 2020 2030 2040 2050 Solving Equations: Examples 1. World population (continued) (x109) t = 50 P3  1950 2.5098 1960 3.0000 P = 8.8185 Average annual rate of change 1970 3.5859 1980 4.2862 1990 5.1233 billion/yr 2000 6.1239 = 72,282,704 people/yr  = 176,369,423 people/yr 2010 7.3199 P2 Average annual rate of change 2020 8.7495 P = 3.6141 2030 10.458 P1  2040 12.500 t = 50 2050 14.942 Section 4.6 v5.0.1

  7. = t log 20 – log 3 + 1960 log 1.018 Solving Equations: Examples 1. World population (continued) When will the population reach 20 billion ? World population P(t) in billions in year t is given by P(t) = 3(1.018)t–1960 20 = 3(1.018)t–1960 Recall that logb xt = t logb x log 20 = log (3 (1.018)t–1960 ) = log 3 + (t – 1960)log (1.018) = 1960 + 106.34 ≈ 2066 … about39years How long till P(t) is 40 billion ? Question: … about39years What is the doubling time for P(t) ? Section 4.6 v5.0.1

  8. { } Solution set: Solving Equations: Examples 2. Solve e2x = e5x–3 ln (e2x) = ln (e5x–3) 2x = 5x – 3 3 = 3x x = 1 3. Solve 2x = –4 Question: What are the domain and range of f(x) = 2x ? Domain = R Range = (0, ) so, 2x > 0 for allx Clearly there is no solution ... OR ... use 1-1 property { 1 } Solution set: Section 4.6 v5.0.1

  9. 74x–3 7 x2 = Zero product rule log7 log7 74x–3 7 x2 = Solving Equations: Examples 4. Solve Since the exponential function is 1-1 then x2 = 4x – 3 x2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 x – 1 = 0 OR x – 3 = 0 x = 1 OR x = 3 NOTE: Could have applied inverse function x2 = 4x – 3 { 1, 3 } Solution set: Question: Could we use log or ln instead of log7 ? Section 4.6 v5.0.1

  10. Solving Equations: Examples 5. Solve 5 ln x = 10 ln x = 2 By definition 2 is the exponent of the base that yields x Thus x = e2≈ (2.71828)2 ≈ 7.38906 6. Solve log3 (1 – x) = 1 By definition 31 = 1 – x Thus x = 1 – 3 = –2 { ~ 7.38906 } Solution set: { –2 } Solution set: Section 4.6 v5.0.1

  11. ln (3 – x) = ) ( x2 – 4 x2 – 4 ln x + 2 x + 2 3 – x = Why not –2 ? Solving Equations: Examples 7. Solve ln (x2 – 4) – ln (x + 2) = ln (3 – x) Using the quotient rule Since the logarithm function is 1-1 x2 – 4 = (x + 2)(3 – x) 2x2 – x – 10 = 0 2x – 5 = 0 OR x + 2 = 0 x = 5/2 OR x = –2 = 3x – x2+ 6 – 2x = (2x – 5)(x + 2) Question: What are the domain and range of the ln function ? { 5/2 } Solution set: Section 4.6 v5.0.1

  12. Think about it ! Section 4.6 v5.0.1

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