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So let's say that the population of rabbits increase according to the law exponential growth.

If a certain population of rabbits has 100 rabbits after the second day and 300 after the fourth day, can we figure out how many rabbits that there were to begin with, or do we not have enough information?

Richy Chai
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  • You do have enough information; see what equations you get from this. – user84413 Aug 28 '16 at 23:14
  • As a hint, assume the function has the form $f(x)=ax^b$. You have two corresponding values of $x$ and $f(x)$, giving you two equations with which to solve for $a$ and $b$. – Kajelad Aug 28 '16 at 23:16

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Let us make the problem more general, knowing that at time $t_1$ the population is $P_1$ and that at time $t_2$ the population is $P_2$.

The first thing to decide is wich form of the exponential model you want to use.

Suppose that you select, just as Kajelad commented $P_t=a t^b$. So, the equations are $$P_1=a t_1^b \qquad , \qquad P_2=a t_2^b$$ Making the ratio, you have $$\frac{P_2}{P_1}=\frac{a t_2^b}{a t_1^b}=\left(\frac{t_2}{t_1}\right)^b$$ Taking logarithms $$\log\left(\frac{P_2}{P_1}\right)=b\log\left(\frac{t_2}{t_1}\right)$$ then $b$. Now, use the first equation to get $a=\frac{P_1}{t_1^b}$.

Another (equivalent) model could be $P_t=a e^{b t}$. Doing the same $$\frac{P_2}{P_1}=\frac{a e^{bt_2}}{a e^{bt_1}}=\frac{ e^{bt_2}}{ e^{bt_1}}=e^{b(t_2-t_1)}$$ Taking logarithms $$\log\left(\frac{P_2}{P_1}\right)=b(t_2-t_1)$$ then $b$. Now, use the first equation to get $a={P_1}{e^{-bt_1}}$.

Now, for the initial population (that is to say at time $t=0$), whatever the choosen model, the population is $a$.

Claude Leibovici
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