Question

Show all work below to answer the following for a certain city that had a population of 150,000 in the year 2010. In 2020, the population was 185,000.
a. Find the constant rate of continuous growth, k, for this population. (Please round to 5 decimal places)

b. When will the population reach 250,000 people?

Answer 1

The constant rate of continuous growth, k, for this population is equal to 2.11935%. And the population  will reach 250,000 people in 24.36 years.

For solving this question, you should apply the Population Growth Equation.

Population Growth Equation

The formula for the Population Growth Equation is:

      $P_f=P_o*(1+\frac{R}{100} )^t$        

Pf= future population

Po=initial population

r=growth rate

t= time (years)

STEP 1 - Find the constant rate of continuous growth, k, for this population.

For this exercise, you have:

Pf= future population= 185,000 in 2020.

Po=initial population =150,000 in 2010.

r=growth rate= ?

t= time (years)=2020-2010=10

Then,

$P_f=P_o*(1+\frac{R}{100} )^t\\ \\ 185000=150000\cdot \left(1+\frac{R}{100}\:\right)^{10}\\ \\ \left(1+\frac{R}{100}\right)^{10}=\frac{185000}{150000} \\ \\ \left(1+\frac{R}{100}\right)^{10}=\frac{37}{30}\\ \\ R=100\sqrt[10]{\frac{37}{30}}-100=2.11935\%$

STEP 2 - Find the t  for population 250,000 people.

$P_f=P_o*(1+\frac{R}{100} )^t\\ \\ 250000=150000\cdot \left(1+\frac{2.11935}{100}\:\right)^{10}\\ \\ \left(1+\frac{2.11935}{100}\right)^{10}=\frac{250000}{150000} \\ \\ \left(1+\frac{2.11935}{100}\right)^t=\frac{5}{3}\\ \\ t\ln \left(1+\frac{2.11935}{100}\right)=\ln \left(\frac{5}{3}\right)\\ \\ t=\frac{\ln \left(\frac{5}{3}\right)}{\ln \left(\frac{102.11935}{100}\right)}\\ \\ t=24.36$

Read more about the population growth equation here:

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