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Solving for y in the equation ln(y) - ln(500-y)=500kt + C

Solving for y in the equation ln(y) - ln(500-y)=500kt + C

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Solving for y in the equation ln(y) - ln(500-y)=500kt + C

(c) (5 points) To overcome issues like those you discussed in part (b), people often use logistic models in epidemiology. These models lead to equations like the following (you will see where this comes from when you take MATH225): In (y) – In (500 – y) = 500kt + C Here y is the number of infected individuals in an epidemic, t is time, k is a rate constant, and C is another constant. Use log and exponent rules to show that 500D y(t) = D+e-500kt C When you encounter an e, re-write it as D to declutter. This is a heavy duty alge- bra problem. If you can do this, you can handle most of the algebraic manipulations that will appear in this class! Hint: the last step is to multiply the numerator and denominator by e -500kt
Transcribed Image Text:(c) (5 points) To overcome issues like those you discussed in part (b), people often use logistic models in epidemiology. These models lead to equations like the following (you will see where this comes from when you take MATH225): In (y) – In (500 – y) = 500kt + C Here y is the number of infected individuals in an epidemic, t is time, k is a rate constant, and C is another constant. Use log and exponent rules to show that 500D y(t) = D+e-500kt C When you encounter an e, re-write it as D to declutter. This is a heavy duty alge- bra problem. If you can do this, you can handle most of the algebraic manipulations that will appear in this class! Hint: the last step is to multiply the numerator and denominator by e -500kt
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