To answer the questions based on the given population model formula [tex]\( A = 24.2 e^{0.0788 t} \)[/tex], let's go through each part step by step.
### Part (a)
The population model is given by:
[tex]\[ A = 24.2 e^{0.0788 t} \][/tex]
where [tex]\( A \)[/tex] is the population in millions and [tex]\( t \)[/tex] is the number of years after 2000.
In 2000, [tex]\( t = 0 \)[/tex].
Substituting [tex]\( t = 0 \)[/tex] into the formula, we get:
[tex]\[ A = 24.2 e^{0.0788 \cdot 0} \][/tex]
Next, calculate the exponent:
[tex]\[ 0.0788 \cdot 0 = 0 \][/tex]
Therefore, the equation simplifies to:
[tex]\[ A = 24.2 e^0 \][/tex]
Since [tex]\( e^0 = 1 \)[/tex]:
[tex]\[ A = 24.2 \cdot 1 \][/tex]
[tex]\[ A = 24.2 \][/tex]
Thus, in 2000, the population of the state was 24.2 million.
### Part (b)
We need to determine when the population of the state will reach 31.4 million. This means we need to solve for [tex]\( t \)[/tex] when [tex]\( A = 31.4 \)[/tex].
Starting with the formula:
[tex]\[ 31.4 = 24.2 e^{0.0788 t} \][/tex]
First, isolate the exponential term by dividing both sides of the equation by 24.2:
[tex]\[ \frac{31.4}{24.2} = e^{0.0788 t} \][/tex]
[tex]\[ 1.2975206611570248 \approx e^{0.0788 t} \][/tex]
To solve for [tex]\( t \)[/tex], take the natural logarithm (ln) of both sides:
[tex]\[ \ln(1.2975206611570248) = \ln(e^{0.0788 t}) \][/tex]
Using the property of logarithms, [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ \ln(1.2975206611570248) = 0.0788 t \][/tex]
Next, solve for [tex]\( t \)[/tex] by dividing both sides by 0.0788:
[tex]\[ t = \frac{\ln(1.2975206611570248)}{0.0788} \][/tex]
Evaluating the division, we obtain the approximate number of years:
[tex]\[ t \approx 3.3052697938016107 \][/tex]
Thus, it will take approximately 3.31 years for the population of the state to reach 31.4 million.
