Question

the formula $a = 23.1e^{0.0151t}$ models the population of a us state, $a$, in millions, $t$ years after 2000.

a. what was the population of the state in 2000?

b. when will the population of the state reach 28.3 million?

a. in 2000, the population of the state was \\(\square\\) million.

b. the population of the state will reach 28.3 million in the year \\(\square\\).
(round to the nearest year as needed.)

Explanation:

Step1: Solve part a

In 2000, \( t = 0 \) (since \( t \) is years after 2000). Substitute \( t = 0 \) into the formula \( A = 23.1e^{0.0151t} \).
We know that \( e^0 = 1 \), so \( A = 23.1e^{0.0151\times0}=23.1\times1 = 23.1 \).

Step2: Solve part b

We need to find \( t \) when \( A = 28.3 \). So set up the equation \( 28.3 = 23.1e^{0.0151t} \).
First, divide both sides by 23.1: \( \frac{28.3}{23.1}=e^{0.0151t} \).
Calculate \( \frac{28.3}{23.1}\approx1.2251 \). So \( 1.2251 = e^{0.0151t} \).
Take the natural logarithm of both sides: \( \ln(1.2251)=\ln(e^{0.0151t}) \).
Since \( \ln(e^x)=x \), we have \( \ln(1.2251)=0.0151t \).
Now, solve for \( t \): \( t=\frac{\ln(1.2251)}{0.0151} \).
Calculate \( \ln(1.2251)\approx0.2027 \), then \( t=\frac{0.2027}{0.0151}\approx13.42 \).
Since \( t \) is years after 2000, the year is \( 2000 + 13 = 2013 \) (rounded to the nearest year).

Answer:

a. \( 23.1 \)
b. \( 2013 \)