Question

The expected pop of a certain bird grows according to the model p(t)=(500)/(1+83.33e^-0.162t)
When will the population reach 300?

Answer 1

you want to solve:

\[p(t) = \dfrac{500}{1+83.33e^{-0.162t}} \color{green}{=300} \]

that means you have to say that:

\[\dfrac{500}{300} =1+83.33e^{-0.162t} \]

Such that:

\[83.33~ e^{-0.162t} = \dfrac{500}{300} - 1 \]

and so on..... :-))

So process the algebra and take some logs.

Answer 2

@IrishBoy123 how come the 500 just went over to the other side how'd you know that

Answer 3

You can do it many ways:

\[\dfrac{500}{1+83.33e^{-0.162t}} \color{green}{=300}\]


\[\dfrac{1+83.33e^{-0.162t}}{500} =\dfrac{1}{300}\]

\[1+83.33e^{-0.162t} =\dfrac{500}{300}\]

\[83.33e^{-0.162t} =\dfrac{500}{300} - 1\]

\[e^{-0.162t} =\dfrac{1}{83.33} \left( \dfrac{500}{300} - 1 \right)\]

\[-0.162t =\ln \left( \dfrac{1}{83.33} \left( \dfrac{500}{300} - 1 \right) \right)\]

Answer 4

@IrishBoy123 I'm really confused on how to get those possibilities ^ I can't seem to even get 1

Answer 5

post your own approach !


we can work through that

people approach these things in different ways. Most of the times, i reckon :-)

Answer 6

how do i write the exponents?

Answer 7

that last thing i posted is merely a question of your posting some numbers into a calculator

This

\[-0.162t =\ln \left( \dfrac{1}{83.33} \left( \dfrac{500}{300} - 1 \right) \right)\]

Followed by:

\[t =-\dfrac{1}{0.162} \ln \left( \dfrac{1}{83.33} \left( \dfrac{500}{300} - 1 \right) \right)\]


there could easily be a typo in there as I have just latexed this on OS :))