The population of a certain country can be modeled by the equation P(t) = 40e^0.02t , where P is the population in millions and t is the number of years since 1900. When will the population be 100 million, 200 million, and 400 million? What do you notice about these time periods?
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100 = 40 * e^(0.02 * t)
5/2 = e^(t/50)
ln(5/2) = t/50
50 * ln(5/2) = t
200 = 40 * e^(0.02 * t)
5 = e^(t/50)
ln(5) = t/50
50 * ln(5) = t
400 = 40 * e^(t/50)
10 = e^(t/50)
ln(10) = t/50
50 * ln(10) = t
45.814536593707753259176360588401
80.471895621705018730037966661309
115.12925464970228420089957273422
There's a common difference between each term.
50 * (ln(10) - ln(5)) = 50 * (ln(5) - ln(5/2))
ln(10) - ln(5) = ln(5) - ln(5/2)
ln(10) + ln(5/2) = ln(5) + ln(5)
ln(25) = ln(25)
That means that it'll be another 50 * ln(2) years before the population reaches 800 million, and another 50 * ln(2) years before it reaches 1.6 billion
5/2 = e^(t/50)
ln(5/2) = t/50
50 * ln(5/2) = t
200 = 40 * e^(0.02 * t)
5 = e^(t/50)
ln(5) = t/50
50 * ln(5) = t
400 = 40 * e^(t/50)
10 = e^(t/50)
ln(10) = t/50
50 * ln(10) = t
45.814536593707753259176360588401
80.471895621705018730037966661309
115.12925464970228420089957273422
There's a common difference between each term.
50 * (ln(10) - ln(5)) = 50 * (ln(5) - ln(5/2))
ln(10) - ln(5) = ln(5) - ln(5/2)
ln(10) + ln(5/2) = ln(5) + ln(5)
ln(25) = ln(25)
That means that it'll be another 50 * ln(2) years before the population reaches 800 million, and another 50 * ln(2) years before it reaches 1.6 billion
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40e^0.02t = 100million
e^0.02t = 2.5million
lne^0.02t = ln(2.5million)
0.02t = 14.7318
t = 14.7318/0.02
t = 736.59
repeat for other population values
e^0.02t = 2.5million
lne^0.02t = ln(2.5million)
0.02t = 14.7318
t = 14.7318/0.02
t = 736.59
repeat for other population values
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hint:
p=100=40e^(0.02t) then solve for t
p=100=40e^(0.02t) then solve for t