A population grows according to a logistic model with a carrying capacity of 10000. An initial population of 100 grows to 1000 in 100 hours. How long will it take for an initial population of 100 to grow to 9000.
SOLUTION :
Growth model is :
P = Po e^(kt)
=> P/Po = e^(kt)
=> 1000/100 = e^(k * 100)
=> 10 = e^(100 k)
Taking natural logarithm :
=> ln(10) = 100 k
=> k = ln(10)/100
When P = 9000.
=> P/Po = 9000/100 = 90
Let time taken be t hours for growing to 9000 from 100.
So,
P/Po = e^(ln(10)/ 100 * t)
=> 90 = e^(ln(10)/100 * t)
Taking ln :
=> ln(90) = ln(10)/100 * t
=> t = 100 ln(90) / ln(10) = 195.42 hours
To grow to 9000 from 100, it will take = 195 .42 hours (ANSWER).
A population grows according to a logistic model with a carrying capacity of 10000. An initial population of 100 grows to 1000 in 100 hours. How long will it take for an initial population of 100 to grow to 9000.
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