solve b and c The populations P in thousands of a certain ci

solve (b) and (c)

The populations P (in thousands) of a certain city from 2005 through 2010 can be modeled by P = 318.2e^kt, where t represents the year, with t = 5 corresponding to 2005. In 2006, the population of the city was about 366,000. (a) Find the value of k. k = Is the population increasing or decreasing? Explain. Because k is negative, the population is decreasing. Because k is negative, the population is increasing. Because k is positive, the population is decreasing. Because k is positive, the population is increasing. (b) Use the model to predict the populations of the city in 2015 and 2020. 2015 P = thousand 2020 P = thousand (c) According to the model, during what year will the population reach 530, 000?

Solution

P= 318.2e^kt

At t=6 P=366000

366000/318.2 =e^6k

1150.21 =e^6k

k=1/6(ln(1150.21)

k=1.1746

Because k is positive population is increasing

In 2015 t=15

P=318.2e^(17.619)

=44857439.1

In 2020 t=20

P=318.2e^(23.492)

=1.59384458 *10^10

C) 530000 =318.2e^(1.1746t)

1/(1.1746) ln(1665.619)

=6.13

So it will be in 2006-2007

Thank you