Explanation Proof: The given logistic growth function is, Q ( t ) = A 1 + B e − k t (1) Substitute Q 1 for Q ( t ) and t 1 for t in the above equation, Q 1 = A 1 + B e − k t 1 1 + B e − k t 1 = A Q 1 B e − k t 1 = A Q 1 − 1 B e − k t 1 = A − Q 1 Q 1 (2) Again, substitute Q 2 for Q ( t ) and t 2 for t in the above equation, Q 2 = A 1 + B e − k t 2 1 + B e − k t 2 = A Q 2 B e − k t 2 = A Q 2 − 1 B e − k t 2 = A − Q 2 Q 2 (3) Divide the equation (2) and equation (3), B e − k t 1 B e − k t 2 = ( A − Q 1 Q 1 ) ( A − Q 2 Q 2 ) e − k t 1 e − k t 2 = ( A − Q 1 Q

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 5.6, Problem 36E

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To show: The value of k is 1t2−t1ln[Q2(A−Q1)Q1(A−Q2)] in the given expression.

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Chapter 5.6, Problem 35E

Chapter 5.6, Problem 37E

Chapter 5 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

Ch. 5.1 - Prob. 1CQ Ch. 5.1 - Prob. 2CQ Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E

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Prob. 24E Ch. 5.5 - Prob. 25E Ch. 5.5 - Prob. 26E Ch. 5.5 - Prob. 27E Ch. 5.5 - find the derivative of the function. 28...Ch. 5.5 - Prob. 29E Ch. 5.5 - Prob. 30E Ch. 5.5 - Prob. 31E Ch. 5.5 - Prob. 32E Ch. 5.5 - Prob. 33E Ch. 5.5 - Prob. 34E Ch. 5.5 - Prob. 35E Ch. 5.5 - Prob. 36E Ch. 5.5 - Prob. 37E Ch. 5.5 - Prob. 38E Ch. 5.5 - Prob. 39E Ch. 5.5 - Prob. 40E Ch. 5.5 - Prob. 41E Ch. 5.5 - Prob. 42E Ch. 5.5 - Prob. 43E Ch. 5.5 - Prob. 44E Ch. 5.5 - Prob. 45E Ch. 5.5 - Prob. 46E Ch. 5.5 - Prob. 47E Ch. 5.5 - Prob. 48E Ch. 5.5 - Prob. 49E Ch. 5.5 - Prob. 50E Ch. 5.5 - Prob. 51E Ch. 5.5 - Prob. 52E Ch. 5.5 - Prob. 53E Ch. 5.5 - Prob. 54E Ch. 5.5 - Prob. 55E Ch. 5.5 - Prob. 56E Ch. 5.5 - Prob. 57E Ch. 5.5 - Prob. 58E Ch. 5.5 - Prob. 59E Ch. 5.5 - Prob. 60E Ch. 5.5 - Prob. 61E Ch. 5.5 - Prob. 62E Ch. 5.5 - Prob. 63E Ch. 5.5 - Prob. 64E Ch. 5.5 - Prob. 65E Ch. 5.5 - Prob. 66E Ch. 5.5 - Prob. 67E Ch. 5.5 - Prob. 68E Ch. 5.5 - Prob. 69E Ch. 5.5 - Prob. 70E Ch. 5.5 - Prob. 71E Ch. 5.5 - Prob. 72E Ch. 5.5 - 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Prob. 17E Ch. 5.6 - Prob. 18E Ch. 5.6 - DEMAND FOR COMPUTERS Universal Instruments found...Ch. 5.6 - Prob. 20E Ch. 5.6 - Prob. 21E Ch. 5.6 - Prob. 22E Ch. 5.6 - GROWTH OF A FRUIT FLY POPULATION On the basis of...Ch. 5.6 - Prob. 24E Ch. 5.6 - Prob. 25E Ch. 5.6 - Prob. 26E Ch. 5.6 - Prob. 27E Ch. 5.6 - Prob. 28E Ch. 5.6 - Prob. 29E Ch. 5.6 - Prob. 30E Ch. 5.6 - Prob. 31E Ch. 5.6 - Prob. 32E Ch. 5.6 - Prob. 33E Ch. 5.6 - Prob. 34E Ch. 5.6 - Prob. 35E Ch. 5.6 - Prob. 36E Ch. 5.6 - Prob. 37E Ch. 5.6 - Prob. 38E Ch. 5.6 - Prob. 1TE Ch. 5.6 - Prob. 2TE Ch. 5.6 - Prob. 3TE Ch. 5.6 - Prob. 4TE Ch. 5.6 - Prob. 5TE Ch. 5.6 - Prob. 6TE Ch. 5.6 - Prob. 7TE Ch. 5.6 - Prob. 8TE Ch. 5 - Prob. 1CRQ Ch. 5 - Prob. 2CRQ Ch. 5 - Prob. 3CRQ Ch. 5 - Prob. 4CRQ Ch. 5 - Prob. 5CRQ Ch. 5 - Prob. 6CRQ Ch. 5 - Prob. 7CRQ Ch. 5 - Prob. 8CRQ Ch. 5 - Prob. 9CRQ Ch. 5 - Prob. 10CRQ Ch. 5 - Prob. 1RE Ch. 5 - Prob. 2RE Ch. 5 - Prob. 3RE Ch. 5 - Prob. 4RE Ch. 5 - Prob. 5RE Ch. 5 - Prob. 6RE Ch. 5 - Prob. 7RE Ch. 5 - Prob. 8RE Ch. 5 - 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The value of k is 1 t 2 − t 1 ln [ Q 2 ( A − Q 1 ) Q 1 ( A − Q 2 ) ] in the given expression.

Solution Summary: The author explains that the given logistic growth function is Q(t)=A1+Be-kt.

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To show: The value of k is 1t2−t1ln[Q2(A−Q1)Q1(A−Q2)] in the given expression.

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Chapter 5 Solutions