Question

café mathematics -- explore
john and leonhard at the café mathematica
john napier is running late in meeting with his friend leonhard euler at a local coffee shop, the café mathematica. john is behind schedule because he has spent all morning making a discovery, namely that:
$f(x) = \log_b x$ has some relation to $x = b^{f(x)}$
while leonhard is waiting for his hopelessly tardy friend, he begins scribbling out the solutions to some exponential problems that have been posed by a few of his professor friends at the local university on some spare napkins at the table.

1. the population of a town increases according to the model

$p(t) = 2500e^{0.0293t}$
where $t$ is the time in years, with $t = 0$ corresponding to 2010.
a. find the projected population of the town in 2012, 2015, and 2018.
b. use a graphing calculator to graph the function for the years 2010 through 2030.
c. use a graphing calculator to approximate the population in 2025 and 2030.
d. verify your answers in part (c) algebraically.

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last updated july 2024
page 10 of 17

Explanation:

Part (a)

Step 1: Determine \( t \) for each year

• For 2012: \( t = 2012 - 2010 = 2 \)
• For 2015: \( t = 2015 - 2010 = 5 \)
• For 2018: \( t = 2018 - 2010 = 8 \)

Step 2: Calculate population for 2012

Use \( P(t) = 2500e^{0.0293t} \), substitute \( t = 2 \):
\( P(2) = 2500e^{0.0293 \times 2} = 2500e^{0.0586} \approx 2500 \times 1.0607 \approx 2651.75 \)

Step 3: Calculate population for 2015

Substitute \( t = 5 \):
\( P(5) = 2500e^{0.0293 \times 5} = 2500e^{0.1465} \approx 2500 \times 1.1580 \approx 2895.00 \)

Step 4: Calculate population for 2018

Substitute \( t = 8 \):
\( P(8) = 2500e^{0.0293 \times 8} = 2500e^{0.2344} \approx 2500 \times 1.2645 \approx 3161.25 \)

Step 1: Determine \( t \) for 2025 and 2030

• 2025: \( t = 2025 - 2010 = 15 \)
• 2030: \( t = 2030 - 2010 = 20 \)

Step 2: Calculate population for 2025

Substitute \( t = 15 \) into \( P(t) = 2500e^{0.0293t} \):
\( P(15) = 2500e^{0.0293 \times 15} = 2500e^{0.4395} \approx 2500 \times 1.5523 \approx 3880.75 \)

Step 3: Calculate population for 2030

Substitute \( t = 20 \) into \( P(t) = 2500e^{0.0293t} \):
\( P(20) = 2500e^{0.0293 \times 20} = 2500e^{0.586} \approx 2500 \times 1.8080 \approx 4520.00 \)

Answer:

• 2012: \( \approx 2652 \)
• 2015: \( \approx 2895 \)
• 2018: \( \approx 3161 \)

Part (d) (Verifying 2025 and 2030 algebraically)