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two bird populations were studied and formulas were developed for each populations growth. the populations p1 and p2 are given by the formulas p1= 30e^kt and p2=60e^0.2t, where i is a constant and t is the time in years since the start of the study. if two populations were equal 10 years after the start of the study what is the approximate value of k? round to two decimals

Respuesta :

Answer:

[tex] 20 e^{k*10} = 60 e^{0.2*10}[/tex]

We can divide both sides by 20 and we got:

[tex] e^{10k}= 3 e^{2}[/tex]

Now we can appply natural log on both sides and we got:

[tex] 10 k = ln(3e^2)[/tex]

Now we can divide both sides of the equation by 10 and we got:

[tex] k = \frac{ln(3e^2)}{10}= 0.30986[/tex]

So then the aproximate value of k is 0.30986 and rounded would be 0.31

Step-by-step explanation:

We have the following two expression:

[tex] p_1 = 20 e^{kt}[/tex]

[tex] p_2 = 60 e^{0.2t}[/tex]

We know that the two populations were equal 10 years after the start of the study, so then we can create the following equation:

[tex] 20 e^{k*10} = 60 e^{0.2*10}[/tex]

We can divide both sides by 20 and we got:

[tex] e^{10k}= 3 e^{2}[/tex]

Now we can appply natural log on both sides and we got:

[tex] 10 k = ln(3e^2)[/tex]

Now we can divide both sides of the equation by 10 and we got:

[tex] k = \frac{ln(3e^2)}{10}= 0.30986[/tex]

So then the aproximate value of k is 0.30986 and rounded would be 0.31

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