Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
We have an exponential growth problem here, it is given by the formula:
[tex]P(t)=P(0)\cdot e^{kt}[/tex]Where t is the time in years, P(0) is the initial population at t=0, and k is a constant.
We already know the population at t=0: P(0)=20,950.
And it doubles every 92 years, then when t=92, P(92)=2*P(0).
If we replace these values we can solve for k as follows:
[tex]\begin{gathered} P(0)=20,950\cdot e^{k\cdot0}=20,950 \\ P(92)=2\cdot20,950=41,900=20,950\cdot e^{k\cdot92} \\ We\text{ can solve for k as follows} \\ 41,900=20,950\cdot e^{k\cdot92} \\ \frac{41,900}{20,950}=e^{k\cdot92} \\ 2=e^{k\cdot92} \\ \text{Apply ln to both sides} \\ \ln 2=\ln e^{k\cdot92} \\ \text{Apply the properties of logarithms:} \\ 0.693=k\cdot92\cdot\ln e \\ \text{Simplify} \\ 0.693=k\cdot92 \\ k=\frac{0.693}{92} \\ k=0.0075 \end{gathered}[/tex]Then, to know the population 460 years from now, we need to replace k=0.0075 and t=460 and solve:
[tex]\begin{gathered} P(460)=20,950\cdot e^{0.0075\cdot460} \\ P(460)=20,950\cdot e^{3.466} \\ P(460)=20,950\cdot32 \\ P(460)=670,400 \end{gathered}[/tex]The population 460 years from now will be 670,400.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
