# Chapter 9 - Section 9.4 - Models for Population Growth - 9.4 Exercises - Page 618: 12

(a) $P(t) = \frac{10,000}{1+24e^{-1.186t}}$ (b) It will take 2.7 years for the population to increase to 5000.

#### Work Step by Step

(a) A logistic equation has this form: $\frac{dP}{dt} = kP(1-\frac{P}{M})$ The solution has this form: $P(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-P_0}{P_0}$ We can find $A$: $A = \frac{M-P_0}{P_0} = \frac{10,000-400}{400} = 24$ We can write the solution: $P(t) = \frac{10,000}{1+24e^{-kt}}$ The population grows to 1200 after 1 year. We can find $k$: $P(t) = \frac{10,000}{1+24e^{-(k)(1)}} = 1200$ $1+24e^{-k} = \frac{10,000}{1200}$ $1+24e^{-k} = 8.33$ $24e^{-k} = 7.33$ $e^{-k} = \frac{7.33}{24}$ $-k = ln(\frac{7.33}{24})$ $k = -ln(\frac{7.33}{24})$ $k = 1.186$ We can write an equation for the population after $t$ years: $P(t) = \frac{10,000}{1+24e^{-1.186t}}$ (b) We can find the time $t$ it takes for the population to increase to 5000: $P(t) = \frac{10,000}{1+24e^{-1.186t}} = 5000$ $1+24e^{-1.186t} = \frac{10,000}{5000}$ $1+24e^{-1.186t} = 2$ $24e^{-1.186t} = 1$ $e^{-1.186t} = \frac{1}{24}$ $-1.186t = ln(\frac{1}{24})$ $t = \frac{1}{-1.186}~ln(\frac{1}{24})$ $t = 2.7$ It will take 2.7 years for the population to increase to 5000.

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