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.