If a population growing exponentially doubles | vedclass

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(A) For exponential growth,the population size at time $t$ is given by the formula $N_t = N_0 e^{rt}$.Given that the population doubles in $3$ years,w

Vedclass · Accepted Answer

$23.1$

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(A) For exponential growth,the population size at time $t$ is given by the formula $N_t = N_0 e^{rt}$.Given that the population doubles in $3$ years,we have $N_t = 2N_0$ when $t = 3$.Substituting these values into the equation: $2N_0 = N_0 e^{r 	imes 3}$.Dividing both sides by $N_0$,we get $2 = e^{3r}$.Taking the natural logarithm $(\ln)$ on both sides: $\ln(2) = 3r$.We know that $\ln(2) \approx 0.693$.So,$0.693 = 3r$.Solving for $r$: $r = 0.693 / 3 = 0.231$.To express this as a percentage: $0.231 	imes 100 = 23.1\%$.

If a population growing exponentially doubles in size in $3$ years,what is the intrinsic rate of increase $(r)$ of the population (in $\%$)? $(e = 2.71828)$

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