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***N= Pe^(k*t) <---this is the equation, not sure how to use it....

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- Thread starter gigi9
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- #1

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***N= Pe^(k*t) <---this is the equation, not sure how to use it....

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HallsofIvy

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Are you sure this doesn't belong in "homework"?

Given N= Pe^{kt}, then since e^{0}= 1, P must be the population for the initial year. You are told that the population increases by 3% each year, so after one year it must be

1.03P: N(t)= 1.03P= Pe^{k(1)}. You can divide by P to get 1.03= e^{k} and so k= ln(1.03).

You can now write the formula as N(t)= P e^{(ln(1.03))t}

but it would be a really good idea to note that e^{kt}= (e^{k})^{t} and since e^{ln(1.03)}= 1.03 the equation is just N(t)= P*(1.03)^{t}.

After 10 years, you have either N(10)= Pe^{(10ln(1.03))} or

N(10)= P*(1.03)^{10}. The "factor" by which it increases is

that number multiplying P: e^{(10ln(1.03))}= (1.03)^{10}.

Notice that that "1.03" is precisely because the population was increasing by 3%= 0.03 each year. If the percentage increase was 100r%, then the factor would be 1+r. To answer the second part, "What percentage increase will double the population every ten years?", solve either Pe^{10ln(1+r)}= 2P or P(1+r)^{10}= 2P for r.

Given N= Pe

1.03P: N(t)= 1.03P= Pe

You can now write the formula as N(t)= P e

but it would be a really good idea to note that e

After 10 years, you have either N(10)= Pe

N(10)= P*(1.03)

that number multiplying P: e

Notice that that "1.03" is precisely because the population was increasing by 3%= 0.03 each year. If the percentage increase was 100r%, then the factor would be 1+r. To answer the second part, "What percentage increase will double the population every ten years?", solve either Pe

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