A lake is stocked with 20 carp. If the carp population grows at a rate of 7.5% per month, when will the pond contain 500 carp? Round your answer to the nearest hundredth.



Answer :

We have a lake with 20 carp and a gowth rate of 7.5% per month, so the equation of the population at time t is:

[tex]\begin{gathered} P(t)=P_0\cdot(1+\frac{7.5}{100})^t \\ \text{Where t is the time in months, P}_0\text{ is the initial population (t=0) and P(t) is the population at time t.} \end{gathered}[/tex]

So, we need to find the value of t when P(t)=500 and P0 (the initial population) is 20:

[tex]\begin{gathered} P(t)=500=20\cdot(1+\frac{7.5}{100})^t \\ \frac{500}{20}=1.075^t \\ \log (25)=\log (1.075^t) \\ \log (25)=t\cdot\log (1.075) \\ t=\frac{\log(25)}{\log(1.075)}=44.51 \end{gathered}[/tex]

The lake contain 500 carp in 44.51 months.