# Lesson 13

Exponential Functions with Base $e$

### Problem 1

The population of a town is growing exponentially and can be modeled by the equation \(f(t) = 42 \boldcdot e^{(0.015t)}\). The population is measured in thousands, and time is measured in years since 1950.

- What was the population of the town in 1950?
- What is the approximate percent increase in the population each year?
- According to this model, approximately what was the population in 1960?

### Solution

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### Problem 2

The revenue of a technology company, in thousands of dollars, can be modeled with an exponential function whose starting value is $395,000 where time \(t\) is measured in years after 2010.

Which function predicts exactly 1.2% of annual growth: \(R(t) = 395 \boldcdot e^{(0.012t)}\) or \(S(t) = 395 \boldcdot (1.012)^t\)? Explain your reasoning.

### Solution

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### Problem 3

How are the functions \(f\) and \(g\) given by \(f(x) = (1.05)^x\) and \(g(x) = e^{0.05x}\) similar? How are they different?

### Solution

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### Problem 4

- A bond is worth $100 and grows in value by 4 percent each year. Explain why the value of the bond after \(t\) years is given by \(100 \boldcdot 1.04^t\).
- A second bond is worth $100 and grows in value by 2 percent each half year. Explain why the value of the bond after \(t\) years is given by \(100 \boldcdot (1.02)^{2t}\).
- A third bond is worth $100 and grows in value by 4 percent each year, but the interest is applied continuously, at every moment. The value of this bond after \(t\) years is given by \(100 \boldcdot e^{(0.04t)}\). Order the bonds from slowest growing to fastest growing. Explain how you know.

### Solution

### Problem 5

The population of a country is growing exponentially, doubling every 50 years. What is the annual growth rate? Explain or show your reasoning.

### Solution

### Problem 6

Which expression has a greater value: \(\log_3 \frac13\) or \(\log_b \frac1b\)? Explain how you know.

### Solution

### Problem 7

The expression \(5 \boldcdot \left(\frac12\right)^d\) models the amount of a radioactive substance, in nanograms, in a sample over time in decades, \(d\). (1 nanogram is a billionth or \(1 \times 10^{\text-9}\) gram.)

- What do the 5 and the \(\frac12\) tell us in this situation?
- When will the sample have less than 0.5 nanogram of the radioactive substance? Express your answer to the nearest half decade. Show your reasoning.
- Show that only about 5 picograms of the substance will remain one century after the sample is measured. (A picogram is a trillionth or \(1 \times 10^{\text-12}\) gram.)

### Solution

### Problem 8

Select **all** true statements about the number \(e\).

\(e\) is a rational number.

\(e\) is approximately 2.718.

\(e\) is an irrational number.

\(e\) is between \(\pi\) and \(\sqrt2\) on the number line.

\(e\) is exactly 2.718.