Question

the population of a city doubled from 1950 to 1988, going from 2.5 million to 5 million people. using the exponential model, $p = p_0e^{rt}$, where $p$ is population, $p_0$ is initial population, and $t$ is time in years, find the annual growth rate $r$ for that period. although the annual growth rate has declined slightly to 1.52% annually, the population of the city is still growing at a tremendous rate. using the initial population of 5 million in 1988 and an annual rate of 1.52%, estimate the population of the city in the year 2010.

the annual growth, $r$, between 1950 and 1988 was \\(\square\\)%
(round to two decimal places as needed.)

Explanation:

Step1: Identify known values

We know that \( P_0 = 2.5 \) million (initial population in 1950), \( P = 5 \) million (population in 1988), and the time \( t = 1988 - 1950 = 38 \) years. The exponential model is \( P = P_0 e^{rt} \).

Step2: Substitute values into the formula

Substitute \( P = 5 \), \( P_0 = 2.5 \), and \( t = 38 \) into \( P = P_0 e^{rt} \):

$$ 5 = 2.5 e^{38r} $$

Step3: Solve for \( r \)

First, divide both sides by \( 2.5 \):

$$ \frac{5}{2.5} = e^{38r} $$

$$ 2 = e^{38r} $$

Then, take the natural logarithm of both sides:

$$ \ln(2) = \ln(e^{38r}) $$

Since \( \ln(e^x) = x \), we have:

$$ \ln(2) = 38r $$

Now, solve for \( r \):

$$ r = \frac{\ln(2)}{38} $$

Calculate \( \ln(2) \approx 0.6931 \), so:

$$ r \approx \frac{0.6931}{38} \approx 0.01824 $$

To convert to a percentage, multiply by 100:

$$ r \approx 0.01824 \times 100 = 1.824\% $$

Round to two decimal places: \( r \approx 1.82\% \)

Answer:

1.82