Question

The population P (in thousands) of Tallahassee, Florida, from 2000 through 2014 can be modeled by P = 150.9e^{kt}, where t represents the year, with t

Comparing two groups
ANSWERED
asked 2021-01-10
The population P (in thousands) of Tallahassee, Florida, from 2000 through 2014 can be modeled by \(P = 150.9e^{kt},\) where t represents the year, with \(t = 0\) corresponding to 2000. In 2005, the population of Tallahassee was about 163,075.
(a) Find the value of k. Is the population increasing or decreasing? Explain.
(b) Use the model to predict the populations of Tallahassee in 2020 and 2025. Are the results reasonable? Explain.
(c) According to the model, during what year will the — populates reach 200,000?

Answers (1)

2021-01-11

Step 1
Given:
The population model is given as,
\(P = 150.9e^{kt}\)
a) To find the value of k, multiply ln on both sides.
\(P = 150.9 \times 1000e^{kt}\)
\(163075 = 150.9 \times 1000e^{5k}\)
\(\frac{163075}{150900} =e^{5k}\)
\(\ln \times \frac{163075}{150900} = ln(e^{5k})\)
\(ln \times \frac{163075}{150900} = 5k\)
\(k = \frac{1}{5} (ln \times \frac{163075}{150900})\)
\(k = \frac{1}{5} \times 0.0776\)
\(k = 0.01552\)
\(k > 0\)
Since the value of k is positive. Hence the population is increasing.
Step 2
b) The population in 2020 is,
\(t = 2020 - 2000\)
\(t = 20\)
\(P = 150.9 \times 1000 \times e^{kt}\)
\(P = 150900 \times e^{0.0155(20)}\)
\(P = 150900 (1.3634)\)
\(P = 205740\)
The population in 2025 is,
\(t = 2025 - 2000\)
\(t = 25\)
\(P = 150.9 \times 1000 \times e^{kt}\)
\(P = 150900 \times e^{0.0155(25)}\)
\(P = 150900 \times (1.4622)\)
\(P = 220658\)
By comparing the population in 2005, the population in 2020 and 2025 is very low amount of growth.
Hence the obtained results are not reasonable.
Step 3
c) To find the year at which the population is 200000
\(P = 150.9 \times 1000 \times e^{kt}\)
\(200000 = 150900 \times e^{0.0155t}\)
\(\frac{200000}{150900} = e^{0.0155t}\)
Now taking ln on both sides,
\(\ln \times \frac{200000}{150900} = \ln(e^{0.0155t})\)
\(ln \times \frac{200000}{150900} = 0.0155t\)
\(t = \frac{1}{0.0155} [ln \times \frac{200000}{150900}]\)
\(t = \frac{1}{0.0155} [ln 1.32538]\)
\(t = \frac{1}{0.0155} [0.2817]\)
\(t = 18.151\)
Hence in a time period of t \(\approx 18\) years the population will be 200000.

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