Part A
Plug t = 0 into the function to find the population in the year 2000.
This works because t represents the number of years since 2000.
So t = 0 represents 2000, t = 1 is 2001, t = 2 is 2002, and so on.
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A = 23.1*e^(0.0153*t)
A = 23.1*e^(0.0153*0)
A = 23.1*e^(0)
A = 23.1*1
A = 23.1
Recall that the variable A is the population in millions.
Saying A = 23.1 is saying there are 23.1 million people (ie 23 million, 100 thousand)
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Answer: 23.1 million
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Part B
Replace A with 28.3 and use natural logarithms to isolate t.
We'll use the log rules Ln(e) = 1 and Ln(x^y) = y*Ln(x).
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A = 23.1*e^(0.0153*t)
28.3 = 23.1*e^(0.0153*t)
23.1*e^(0.0153*t) = 28.3
e^(0.0153*t) = 28.3/23.1
e^(0.0153*t) = 1.22510822510822
Ln[ e^(0.0153*t) ] = Ln[ 1.22510822510822 ]
0.0153*t*Ln[ e ] = Ln[ 1.22510822510822 ]
0.0153*t*1 = Ln[ 1.22510822510822 ]
0.0153*t = Ln[ 1.22510822510822 ]
t = Ln[ 1.22510822510822 ]/0.0153
t = 13.2698815112052
It will take approximately 13.2699 years for the population to reach the desired target.
This result is between 13 and 14.
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Answer: Between 2013 and 2014
We want to study the population formula of a given US state.
a) 23.1 millions.
b) in 2013 - 2014
The population state is:
A = 23.1*e^(0.0153*t)
In millions.
Where t represents the number of years after 2000.
a) For the population at the year 2000, we need to evaluate the function at t = 0 (2000 is 0 years after 2000).
A(0) = 23.1*e^(0.0153*0) = 23.1
The population at year 2000 was 23.1 millions.
b) Now we want to solve:
A(t) = 28.3 = 23.1*e^(0.0153*t)
28.3/23.1 = e^(0.0153*t)
1.23 = e^(0.0153*t)
Now we can apply the natural logarithm to both sides:
ln(1.23) = ln( e^(0.0153*t) )
ln(1.23) = 0.0153*t
ln(1.23)/0.0153 = t = 13.5
So 13.5 years after 2000 the population will be 28.3 million.
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