Question

In​ 2012, the population of a city was 6.57 million. The exponential growth rate was 2.87​% per year. ​a) Find the exponential growth function. ​b) Estimate the population of the city in 2018. ​c) When will the population of the city be 10 ​million? ​d) Find the doubling time. ​a) The exponential growth function is ​P(t)equals nothing​, where t is in terms of the number of years since 2012 and​ P(t) is the population in millions

Answer 1

Answer:

a) $P(t) = 6.57e ^{0.0287t}$

b) $P(6) = 7,805\ million$

c) $t = 14.64\ years$

d) $t = 24.15\ years$

Step-by-step explanation:

a) The function of exponential growth for a population has the following formula:

$P(t) = p_0e ^{rt}$

In this equation:

$p_0$ is the initial population

r is the growth rate

t is the time in years

In this problem we know that

$r = 2.87\% = 0.0287$

$p_0= 6.57$ million in the year 2012.

So the equation is:

$P(t) = 6.57e ^{0.0287t}$

Where $t = 0$ represents the year 2012

b) If $t = 0$ in 2012, then in 2018 $t = 6$

The population in 2018 is:

$P(t = 6) = 6.57e ^{0.0287(6)}\\\\P(6) = 7,805\ million$

c) To know when the population is equal to 10 million we must equal P(t) to 10 and solve for t.

$P(t) = 10 = 6.57e ^{0.0287t}\\\\\frac{10}{6.57} = e ^{0.0287t}\\\\ln(\frac{10}{6.57}) = 0.0287t\\\\t = \frac{ln(\frac{10}{6.57})}{0.0287}$

$t = 14.64\ years$

d) The function is doubled when $P(t) = 2p_0$

$P(t) = 2(6.57) = 13.14 = 6.57e ^{0.0287t}$

We solve for t.

$\frac{13.14}{6.57} = e ^{0.0287t}\\\\ln(\frac{13.14}{6.57}) = 0.0287t\\\\t = \frac{ln(\frac{13.14}{6.57})}{0.0287})$

$t = 24.15\ years$