5.1 Exponential Growth and Decay Models
- Recognize and model exponential growth and decay
- Compare linear and exponential growth
- Distinguish between exponential and power functions
Comparing Exponential and Linear Growth
Let's start with a real-world comparison that shows exactly how exponential growth differs from the linear growth you already know.
Why does this matter? Many real-world quantities — population growth, investment returns, viral spread, radioactive decay — follow exponential patterns, not linear ones. Understanding the difference is essential for making accurate predictions in business, science, and everyday life.
Consider two social media sites expanding their user bases:
- Site A has 10,000 users and adds 1,500 new users each month
- Site B has 10,000 users and increases the number of users by \(10\%\) each month
Site A grows by a constant number each month — that's linear growth. If \(x\) is the number of months that have passed and \(y\) is the number of users, then:
\[y = 10000 + 1500x\]Site B grows by a constant percent each month. Growth that occurs at a constant percent per unit of time is called exponential growth.
The table below compares user counts for each site over 12 months. The first 4 months show the detailed calculations:
| Month | Users at Site A | Users at Site B |
|---|---|---|
| 0 | 10,000 | 10,000 |
| 1 | \(10000 + 1500 = 11{,}500\) | \(10000 + 10\%\text{ of }10000 = 10000(1.10) = 11{,}000\) |
| 2 | \(11500 + 1500 = 13{,}000\) | \(11000(1.10) = 12{,}100\) |
| 3 | \(13000 + 1500 = 14{,}500\) | \(12100(1.10) = 13{,}310\) |
| 4 | \(14500 + 1500 = 16{,}000\) | \(13310(1.10) = 14{,}641\) |
| 5 | 17,500 | 16,105 |
| 6 | 19,000 | 17,716 |
| 7 | 20,500 | 19,487 |
| 8 | 22,000 | 21,436 |
| 9 | 23,500 | 23,579 |
| 10 | 25,000 | 25,937 |
| 11 | 26,500 | 28,531 |
| 12 | 28,000 | 31,384 |
Notice that linear growth (Site A) starts out faster, but exponential growth (Site B) eventually overtakes it and grows much faster.
For Site B, let's re-express the calculations to discover the pattern and develop a formula:
\[\text{Month 1: } y = 10000(1.1) = 11{,}000\] \[\text{Month 2: } y = 11000(1.1) = 10000(1.1)(1.1) = \mathbf{10000(1.1)^2} = 12{,}100\] \[\text{Month 3: } y = 12100(1.1) = 10000(1.1)^2(1.1) = \mathbf{10000(1.1)^3} = 13{,}310\] \[\text{Month 4: } y = 13310(1.1) = 10000(1.1)^3(1.1) = \mathbf{10000(1.1)^4} = 14{,}641\]By looking at this pattern, we can generalize: after \(x\) months, the number of users \(y\) is given by:
\[y = 10000(1.1)^x\]Notice the structure: you take the initial value (10,000) and multiply by the growth factor (\(1.1\)) raised to the power of time (\(x\)). This \(y = a \cdot b^x\) pattern is the foundation of every exponential function you'll see in this chapter.
Using Exponential Functions to Model Growth and Decay
In exponential growth, the value of \(y\) increases at a constant percentage rate as \(x\) (or \(t\)) increases. Real-world examples include:
- The number of residents in a city that grows at a constant percent rate
- The amount of money in a bank account earning compound interest (deposited once, no withdrawals)
In exponential decay, the value of \(y\) decreases at a constant percentage rate as \(x\) (or \(t\)) increases. Real-world examples include:
- The value of a car or equipment that depreciates at a constant percent rate
- The amount of a drug remaining in the body after ingestion
- The amount of radioactive material remaining over time
Exponential decay isn't just a math concept — it's how doctors determine drug dosages (how fast medication leaves your body), how archaeologists date ancient artifacts (carbon-14 decay), and how businesses plan equipment replacement schedules.
Exponential functions often model quantities as a function of time, so we frequently use the letter \(t\) as the independent variable instead of \(x\).
| Exponential Growth | Exponential Decay | |
|---|---|---|
| Description | Quantity grows by a constant percent per unit of time | Quantity decreases by a constant percent per unit of time |
| Form | \(y = ab^x\) | \(y = ab^x\) |
| Initial value | \(a > 0\) (value when \(x = 0\)) | \(a > 0\) (value when \(x = 0\)) |
| Base | \(b > 1\) | \(0 < b < 1\) |
| Rate | \(r > 0\) where \(b = 1 + r\) | \(r < 0\) where \(b = 1 + r\) |
In general, the domain of exponential functions is the set of all real numbers. The range is the set of all positive real numbers.
In most applications, the independent variable \(x\) or \(t\) represents time. When it represents time, we often restrict the domain to non-negative values (\(x \geq 0\)) so the application makes sense. When we restrict the domain:
- For exponential growth (\(b > 1\), \(a > 0\)): if \(x \geq 0\), then the range is \(y \geq a\)
- For exponential decay (\(0 < b < 1\), \(a > 0\)): if \(x \geq 0\), then the range is \(0 < y \leq a\)
Consider the growth models for social media sites A and B, where \(x\) = number of months since the site was started and \(y\) = number of users.
The number of users for Site A follows the linear growth model:
\[y = 10000 + 1500x\]The number of users for Site B follows the exponential growth model:
\[y = 10000(1.1^x)\]For each site, use the function to calculate the number of users at the end of the first year, to verify the values in the table. Then use the functions to predict the number of users after 30 months.
Example 5.1.1 Solution
Since \(x\) is measured in months, then \(x = 12\) at the end of one year.
Linear Growth Model:
When \(x = 12\) months: \(y = 10000 + 1500(12) = 28{,}000\) users
When \(x = 30\) months: \(y = 10000 + 1500(30) = 55{,}000\) users
Exponential Growth Model:
When \(x = 12\) months: \(y = 10000(1.1^{12}) = 31{,}384\) users
When \(x = 30\) months: \(y = 10000(1.1^{30}) = 174{,}494\) users
We see that as \(x\) gets larger, the exponential growth function grows much faster than the linear function. Even though the linear function initially grew faster, the exponential function eventually dominates. This is an important characteristic: exponential growth functions always grow faster and larger in the long run than linear growth functions.
It's helpful to use function notation, writing \(y = f(t) = ab^t\), so you can clearly specify the value of \(t\) at which the function is evaluated. For example, \(f(5)\) means "evaluate at \(t = 5\)."
A forest has a population of 2,000 squirrels that is increasing at the rate of \(3\%\) per year. Let \(t\) = number of years and \(y = f(t)\) = number of squirrels at time \(t\).
a. Find the exponential growth function that models the number of squirrels at the end of \(t\) years.
b. Use the function to find the number of squirrels after 5 years and after 10 years.
Example 5.1.2 Solution
a. The exponential growth function is \(y = f(t) = ab^t\), where:
- \(a = 2000\) (the initial population)
- The annual growth rate is \(3\%\) per year, so \(r = 0.03\)
- \(b = 1 + r = 1 + 0.03 = 1.03\)
Answer: The exponential growth function is \(y = f(t) = 2000(1.03^t)\)
b. After 5 years: \(y = f(5) = 2000(1.03^5) \approx 2{,}319\) squirrels
After 10 years: \(y = f(10) = 2000(1.03^{10}) \approx 2{,}688\) squirrels
A large lake has a population of 1,000 frogs. Unfortunately, the frog population is decreasing at the rate of \(5\%\) per year. Let \(t\) = number of years and \(y = g(t)\) = the number of frogs at time \(t\).
a. Find the exponential decay function that models the population of frogs.
b. Calculate the size of the frog population after 10 years.
Example 5.1.3 Solution
a. The exponential decay function is \(y = g(t) = ab^t\), where:
- \(a = 1000\) (the initial population)
- The annual decay rate is \(5\%\) per year. The words "decrease" and "decay" indicate that \(r\) is negative: \(r = -0.05\)
- \(b = 1 + r = 1 + (-0.05) = 0.95\)
Answer: The exponential decay function is \(y = g(t) = 1000(0.95^t)\)
b. After 10 years: \(y = g(10) = 1000(0.95^{10}) \approx 599\) frogs
A population of bacteria is given by the function \(y = f(t) = 100(2^t)\), where \(t\) is time measured in hours and \(y\) is the number of bacteria.
a. What is the initial population?
b. What happens to the population in the first hour?
c. How long does it take for the population to reach 800 bacteria?
Example 5.1.4 Solution
a. The initial population is 100 bacteria. We know this because \(a = 100\) and because at time \(t = 0\):
\[f(0) = 100(2^0) = 100(1) = 100\]b. At the end of 1 hour: \(y = f(1) = 100(2^1) = 100(2) = 200\) bacteria. The population has doubled during the first hour.
c. We need to find the time \(t\) at which \(f(t) = 800\). Substitute 800 for \(y\):
\[y = f(t) = 100(2^t)\] \[800 = 100(2^t)\]Divide both sides by 100 to isolate the exponential expression:
\[8 = 2^t\]Since \(8 = 2^3\), it takes \(t = 3\) hours for the population to reach 800 bacteria.
Two important notes about this example:
- In solving \(8 = 2^t\), we "knew" that \(t = 3\). But we usually cannot determine the value just by looking at the equation. Later, we will use logarithms to solve equations with the variable in the exponent.
- To solve \(800 = 100(2^t)\), we divided both sides by 100 to isolate the exponential expression \(2^t\). We cannot multiply 100 by 2. The exponent applies only to the quantity immediately before it — the exponent \(t\) applies only to the base of 2, not to \(100 \times 2\).
Comparing Linear, Exponential, and Power Functions
To identify the type of function from its formula, you need to carefully note where the variable appears in the formula.
| Type | Form | Key Feature |
|---|---|---|
| Linear | \(y = ax + b\) | Variable \(x\) is in the first degree; \(a\) is slope, \(b\) is \(y\)-intercept |
| Exponential | \(y = ab^x\) | Variable \(x\) is in the exponent; base \(b > 0\) |
| Power | \(y = cx^p\) | Variable \(x\) is in the base; exponent \(p \neq 0\) is a fixed number |
For exponential functions:
- If \(b > 1\), the function represents exponential growth
- If \(0 < b < 1\), the function represents exponential decay
Let's compare three specific functions:
- Linear: \(y = f(x) = 2x\)
- Exponential: \(y = g(x) = 2^x\)
- Power: \(y = h(x) = x^2\)
| \(x\) | \(y = f(x) = 2x\) | \(y = g(x) = 2^x\) | \(y = h(x) = x^2\) |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 1 | 2 | 2 | 1 |
| 2 | 4 | 4 | 4 |
| 3 | 6 | 8 | 9 |
| 4 | 8 | 16 | 16 |
| 5 | 10 | 32 | 25 |
| 6 | 12 | 64 | 36 |
| 10 | 20 | 1,024 | 100 |
Notice the key difference in how each function grows:
- Linear: For equal intervals of change in \(x\), \(y\) increases by a constant amount (always +2)
- Exponential: For equal intervals of change in \(x\), \(y\) increases by a constant ratio (always ×2)
- Power: The growth rate varies — it's not constant by amount or ratio
The exponential function eventually grows larger and faster than both the linear and power functions. This is always true of exponential growth as \(x\) gets large enough.
Classify the functions below as exponential, linear, or power functions.
a. \(y = 10x^3\)
b. \(y = 1000 - 30x\)
c. \(y = 1000(1.05^x)\)
d. \(y = 500(0.75^x)\)
e. \(y = 10\sqrt[3]{x} = 10x^{1/3}\)
f. \(y = 5x - 1\)
g. \(y = 6/x^2 = 6x^{-2}\)
Example 5.1.5 Solution
Exponential functions (variable is in the exponent):
- c. \(y = 1000(1.05^x)\) — base is \(b = 1.05\)
- d. \(y = 500(0.75^x)\) — base is \(b = 0.75\)
Linear functions (variable is in the first degree):
- b. \(y = 1000 - 30x\)
- f. \(y = 5x - 1\)
Power functions (variable is in the base, exponent is a fixed number):
- a. \(y = 10x^3\) — exponent is \(p = 3\)
- e. \(y = 10x^{1/3}\) — exponent is \(p = 1/3\)
- g. \(y = 6x^{-2}\) — exponent is \(p = -2\)
Natural Base: \(e\)
The number \(e\) is often used as the base of an exponential function. It is called the natural base.
\(e\) is an irrational number with an infinite, non-repeating decimal expansion.
Why use this strange number \(e\) instead of a nice round number? The number \(e\) shows up naturally when you model continuous processes — growth that happens every instant rather than at fixed intervals (like monthly or yearly). Section 6.2 shows where \(e\) comes from and why it matters. You'll use \(e\) heavily in Chapter 6 when calculating continuously compounded interest.
When \(e\) is the base in an exponential growth or decay function, we call it continuous growth or continuous decay. Any exponential function can be written in the form:
\[y = ae^{kx}\]where \(k\) is called the continuous growth or decay rate:
- If \(k > 0\), the function represents exponential growth
- If \(k < 0\), the function represents exponential decay
- \(a\) is the initial value
We can convert between the forms \(y = ab^x\) and \(y = ae^{kx}\) using the relationship \(b = e^k\).
In general, if we know one form of the equation, we can find the other. To find \(k\) when we know \(b\), we use the natural logarithm: \(k = \ln b\). We'll cover logarithms later in this chapter.
| \(y = ab^x\) | \(y = a(1+r)^x\) | \(y = ae^{kx}\), \(k \neq 0\) | |
|---|---|---|---|
| Initial value | \(a > 0\) | \(a > 0\) | \(a > 0\) |
| Relationships | \(b > 0\) | \(b = 1 + r\) | \(b = e^k\) and \(k = \ln b\) |
| Growth | \(b > 1\) | \(r > 0\) | \(k > 0\) |
| Decay | \(0 < b < 1\) | \(r < 0\) | \(k < 0\) |
The value of houses in a city are increasing at a continuous growth rate of \(6\%\) per year. For a house that currently costs $400,000:
a. Write the exponential growth function in the form \(y = ae^{kx}\).
b. What would be the value of this house 4 years from now?
c. Rewrite the exponential growth function in the form \(y = ab^x\).
d. Find and interpret \(r\).
Example 5.1.6 Solution
a. The initial value of the house is \(a = 400{,}000\).
The continuous growth rate is \(6\%\) per year, so \(k = 0.06\).
The growth function is: \(y = 400000e^{0.06x}\)
b. After 4 years:
\[y = 400000e^{0.06(4)} = \$508{,}500\]c. To rewrite in \(y = ab^x\) form, use \(b = e^k\):
\[b = e^{0.06} = 1.06183657 \approx 1.0618\] \[y = 400000(1.0618)^x\]d. To find \(r\), use \(b = 1 + r\):
\[1 + r = 1.0618\] \[r = 0.0618\]The value of the house is increasing at an annual rate of \(6.18\%\).
Suppose that the value of a certain model of new car decreases at a continuous decay rate of \(8\%\) per year. For a car that costs $20,000 when new:
a. Write the exponential decay function in the form \(y = ae^{kx}\).
b. What would be the value of this car 5 years from now?
c. Rewrite the exponential decay function in the form \(y = ab^x\).
d. Find and interpret \(r\).
Example 5.1.7 Solution
a. The initial value of the car is \(a = 20{,}000\).
The continuous decay rate is \(8\%\) per year, so \(k = -0.08\).
The decay function is: \(y = 20000e^{-0.08x}\)
b. After 5 years:
\[y = 20000e^{-0.08(5)} = \$13{,}406.40\]c. To rewrite in \(y = ab^x\) form, use \(b = e^k\):
\[b = e^{-0.08} = 0.9231163464 \approx 0.9231\] \[y = 20000(0.9231)^x\]d. To find \(r\), use \(b = 1 + r\):
\[b = 0.9231\] \[1 + r = 0.9231\] \[r = 0.9231 - 1 = -0.0769\]The value of the car is decreasing at an annual rate of \(7.69\%\).
Problem Set 5.1
Problem Set 5.1
Identify each as an exponential, linear, or power function:
1. \(y = 640(1.25^x)\)
Problem 1 Solution
The variable \(x\) is in the exponent and the base is the number \(1.25\).
Answer: Exponential function
2. \(y = 640(x^{1.25})\)
Problem 2 Solution
The variable \(x\) is in the base and the exponent is the fixed number \(1.25\).
Answer: Power function
3. \(y = 640(1.25x)\)
Problem 3 Solution
Rewrite as \(y = 640(1.25)x = 800x\). The variable \(x\) is in the first degree.
Answer: Linear function
4. \(y = 1.05x - 2.5\)
Problem 4 Solution
The variable \(x\) is in the first degree with slope \(1.05\) and \(y\)-intercept \(-2.5\).
Answer: Linear function
5. \(y = 90 - (4/5)x\)
Problem 5 Solution
Rewrite as \(y = -(4/5)x + 90\). The variable \(x\) is in the first degree with slope \(-4/5\) and \(y\)-intercept \(90\).
Answer: Linear function
6. \(y = 42(0.92^x)\)
Problem 6 Solution
The variable \(x\) is in the exponent and the base is the number \(0.92\).
Answer: Exponential function
7. \(y = 37(x^{0.25})\)
Problem 7 Solution
The variable \(x\) is in the base and the exponent is the fixed number \(0.25\).
Answer: Power function
8. \(y = 4(1/3)^x\)
Problem 8 Solution
The variable \(x\) is in the exponent and the base is the number \(1/3\).
Answer: Exponential function
Indicate if the function represents exponential growth or exponential decay:
9. \(y = 127e^{-0.35t}\)
Problem 9 Solution
This is in the form \(y = ae^{kt}\) with \(k = -0.35 < 0\).
Answer: Exponential decay
10. \(y = 70(0.8^t)\)
Problem 10 Solution
This is in the form \(y = ab^t\) with \(b = 0.8\). Since \(0 < b < 1\):
Answer: Exponential decay
11. \(y = 453(1.2^t)\)
Problem 11 Solution
This is in the form \(y = ab^t\) with \(b = 1.2\). Since \(b > 1\):
Answer: Exponential growth
12. \(y = 16e^{0.2t}\)
Problem 12 Solution
This is in the form \(y = ae^{kt}\) with \(k = 0.2 > 0\).
Answer: Exponential growth
In each of the following, \(y\) is an exponential function of \(t\) stated in the form \(y = ae^{kt}\) where \(t\) represents time in years. For each:
a. Re-express each function in the form \(y = ab^t\) (state the value of \(b\) accurate to 4 decimal places)
b. State the annual growth rate or annual decay rate as a percent, accurate to 2 decimal places
13. \(y = 127e^{-0.35t}\)
Problem 13 Solution
Step 1: Find \(b = e^k = e^{-0.35} = 0.7047\)
a. \(y = 127(0.7047)^t\)
Step 2: Find \(r = b - 1 = 0.7047 - 1 = -0.2953\)
b. The annual decay rate is \(-29.53\%\)
Verification: \(e^{-0.35} = 0.704688... \approx 0.7047\) ✓
14. \(y = 16e^{0.4t}\)
Problem 14 Solution
Step 1: Find \(b = e^k = e^{0.4} = 1.4918\)
a. \(y = 16(1.4918)^t\)
Step 2: Find \(r = b - 1 = 1.4918 - 1 = 0.4918\)
b. The annual growth rate is \(49.18\%\)
Verification: \(e^{0.4} = 1.49182... \approx 1.4918\) ✓
15. \(y = 17250e^{0.24t}\)
Problem 15 Solution
Step 1: Find \(b = e^k = e^{0.24} = 1.2712\)
a. \(y = 17250(1.2712)^t\)
Step 2: Find \(r = b - 1 = 1.2712 - 1 = 0.2712\)
b. The annual growth rate is \(27.12\%\)
Verification: \(e^{0.24} = 1.27125... \approx 1.2712\) ✓
16. \(y = 4700e^{-0.07t}\)
Problem 16 Solution
Step 1: Find \(b = e^k = e^{-0.07} = 0.9324\)
a. \(y = 4700(0.9324)^t\)
Step 2: Find \(r = b - 1 = 0.9324 - 1 = -0.0676\)
b. The annual decay rate is \(-6.76\%\)
Verification: \(e^{-0.07} = 0.93239... \approx 0.9324\) ✓
Identify if the function represents exponential growth, exponential decay, linear growth, or linear decay. In each case, write the function and find the value at the indicated time.
17. A house was purchased for $350,000 in the year 2010. The value has been increasing by $7,000 per year. Write the function and find the value of the house after 5 years.
Problem 17 Solution
Step 1: Identify the type. The value increases by a constant amount ($7,000/year), so this is linear growth.
Step 2: Write the function. Initial value \(a = 350{,}000\), slope \(m = 7{,}000\):
\[y = 350000 + 7000t\]Step 3: Find value after 5 years (\(t = 5\)):
\[y = 350000 + 7000(5) = 350000 + 35000 = 385{,}000\]Answer: Linear growth; \(y = 350000 + 7000t\); after 5 years the house is worth $385,000.
Verification: \(350000 + 35000 = 385000\) ✓
18. A house was purchased for $350,000 in the year 2010. The value has been increasing at the rate of 2% per year. Write the function and find the value of the house after 5 years.
Problem 18 Solution
Step 1: Identify the type. The value increases by a constant percent (2%/year), so this is exponential growth.
Step 2: Write the function. \(a = 350{,}000\), \(r = 0.02\), \(b = 1 + 0.02 = 1.02\):
\[y = 350000(1.02)^t\]Step 3: Find value after 5 years (\(t = 5\)):
\[y = 350000(1.02)^5 = 350000(1.10408...) \approx 386{,}428\]Answer: Exponential growth; \(y = 350000(1.02)^t\); after 5 years the house is worth approximately $386,428.
Verification: \(1.02^5 = 1.10408...\), \(350000 \times 1.10408 = 386428\) ✓
19. A lab purchases new equipment for $50,000. Its value depreciates over time. The value decreases at the rate of 6% annually. Write the function and find the value after 10 years.
Problem 19 Solution
Step 1: Identify the type. The value decreases by a constant percent (6%/year), so this is exponential decay.
Step 2: Write the function. \(a = 50{,}000\), \(r = -0.06\), \(b = 1 + (-0.06) = 0.94\):
\[y = 50000(0.94)^t\]Step 3: Find value after 10 years (\(t = 10\)):
\[y = 50000(0.94)^{10} = 50000(0.53862...) \approx 26{,}931\]Answer: Exponential decay; \(y = 50000(0.94)^t\); after 10 years the equipment is worth approximately $26,931.
Verification: \(0.94^{10} = 0.53862...\), \(50000 \times 0.53862 = 26931\) ✓
20. A lab purchases new equipment for $50,000. Its value depreciates over time. The value decreases by $3,000 annually. Write the function and find the value after 10 years.
Problem 20 Solution
Step 1: Identify the type. The value decreases by a constant amount ($3,000/year), so this is linear decay.
Step 2: Write the function. \(a = 50{,}000\), slope \(= -3{,}000\):
\[y = 50000 - 3000t\]Step 3: Find value after 10 years (\(t = 10\)):
\[y = 50000 - 3000(10) = 50000 - 30000 = 20{,}000\]Answer: Linear decay; \(y = 50000 - 3000t\); after 10 years the equipment is worth $20,000.
Verification: \(50000 - 30000 = 20000\) ✓
21. A population of bats in a cave has 200 bats. The population is increasing by 10 bats annually. Write the function. How many bats live in the cave after 7 years?
Problem 21 Solution
Step 1: Identify the type. The population increases by a constant amount (10 bats/year), so this is linear growth.
Step 2: Write the function. \(a = 200\), slope \(= 10\):
\[y = 200 + 10t\]Step 3: Find population after 7 years (\(t = 7\)):
\[y = 200 + 10(7) = 200 + 70 = 270\]Answer: Linear growth; \(y = 200 + 10t\); after 7 years there are 270 bats.
Verification: \(200 + 70 = 270\) ✓
22. A population of bats in a cave has 200 bats. The population is increasing at the rate of 5% annually. Write the function. How many bats live in the cave after 7 years?
Problem 22 Solution
Step 1: Identify the type. The population increases by a constant percent (5%/year), so this is exponential growth.
Step 2: Write the function. \(a = 200\), \(r = 0.05\), \(b = 1.05\):
\[y = 200(1.05)^t\]Step 3: Find population after 7 years (\(t = 7\)):
\[y = 200(1.05)^7 = 200(1.40710...) \approx 281\]Answer: Exponential growth; \(y = 200(1.05)^t\); after 7 years there are approximately 281 bats.
Verification: \(1.05^7 = 1.40710...\), \(200 \times 1.40710 = 281.4 \approx 281\) ✓
23. A population of a certain species of bird in a state park has 300 birds. The population is decreasing at the rate of 7% per year. Write the function. How many birds are in the population after 6 years?
Problem 23 Solution
Step 1: Identify the type. The population decreases by a constant percent (7%/year), so this is exponential decay.
Step 2: Write the function. \(a = 300\), \(r = -0.07\), \(b = 1 + (-0.07) = 0.93\):
\[y = 300(0.93)^t\]Step 3: Find population after 6 years (\(t = 6\)):
\[y = 300(0.93)^6 = 300(0.64716...) \approx 194\]Answer: Exponential decay; \(y = 300(0.93)^t\); after 6 years there are approximately 194 birds.
Verification: \(0.93^6 = 0.64716...\), \(300 \times 0.64716 = 194.1 \approx 194\) ✓
24. A population of a certain species of bird in a state park has 300 birds. The population decreases by 20 birds per year. Write the function. How many birds are in the population after 6 years?
Problem 24 Solution
Step 1: Identify the type. The population decreases by a constant amount (20 birds/year), so this is linear decay.
Step 2: Write the function. \(a = 300\), slope \(= -20\):
\[y = 300 - 20t\]Step 3: Find population after 6 years (\(t = 6\)):
\[y = 300 - 20(6) = 300 - 120 = 180\]Answer: Linear decay; \(y = 300 - 20t\); after 6 years there are 180 birds.
Verification: \(300 - 120 = 180\) ✓
In problems 25–28, the problem represents exponential growth or decay and states the CONTINUOUS growth rate or continuous decay rate. Write the exponential growth or decay function and find the value at the indicated time.
Hint: Use the form of the exponential function that is appropriate when the CONTINUOUS growth or decay rate is given.
25. A population of 400 microbes increases at the continuous growth rate of \(26\%\) per day. Write the function and find the number of microbes in the population at the end of 7 days.
Problem 25 Solution
Step 1: Since the problem states a continuous growth rate, use the form \(y = ae^{kt}\).
\(a = 400\), \(k = 0.26\) (continuous growth rate in decimal form)
\[y = 400e^{0.26t}\]Step 2: Find the population after 7 days (\(t = 7\)):
\[y = 400e^{0.26(7)} = 400e^{1.82} = 400(6.17253...) \approx 2{,}469\]Answer: \(y = 400e^{0.26t}\); after 7 days there are approximately 2,469 microbes.
Verification: \(e^{1.82} = 6.17253...\), \(400 \times 6.17253 = 2469\) ✓
26. The price of a machine needed by a production factory is $28,000. The business expects to replace the machine in 4 years. Due to inflation the price of the machine is increasing at the continuous rate of \(3.5\%\) per year. Write the function and find the value of the machine 4 years from now.
Problem 26 Solution
Step 1: Since the problem states a continuous growth rate, use \(y = ae^{kt}\).
\(a = 28{,}000\), \(k = 0.035\)
\[y = 28000e^{0.035t}\]Step 2: Find the price after 4 years (\(t = 4\)):
\[y = 28000e^{0.035(4)} = 28000e^{0.14} = 28000(1.15027...) \approx 32{,}208\]Answer: \(y = 28000e^{0.035t}\); after 4 years the machine will cost approximately $32,208.
Verification: \(e^{0.14} = 1.15027...\), \(28000 \times 1.15027 = 32208\) ✓
27. A population of an endangered species consists of 4,000 animals of that species. The population is decreasing at the continuous rate of \(12\%\) per year. Write the function and find the size of the population at the end of 10 years.
Problem 27 Solution
Step 1: Since the problem states a continuous decay rate, use \(y = ae^{kt}\).
\(a = 4{,}000\), \(k = -0.12\) (decay, so \(k\) is negative)
\[y = 4000e^{-0.12t}\]Step 2: Find the population after 10 years (\(t = 10\)):
\[y = 4000e^{-0.12(10)} = 4000e^{-1.2} = 4000(0.30119...) \approx 1{,}205\]Answer: \(y = 4000e^{-0.12t}\); after 10 years there are approximately 1,205 animals.
Verification: \(e^{-1.2} = 0.30119...\), \(4000 \times 0.30119 = 1204.8 \approx 1205\) ✓
28. A business buys a computer system for $12,000. The value of the system is depreciating and decreases at the continuous rate of 20% per year. Write the function and find the value at the end of 3 years.
Problem 28 Solution
Step 1: Since the problem states a continuous decay rate, use \(y = ae^{kt}\).
\(a = 12{,}000\), \(k = -0.20\) (decay, so \(k\) is negative)
\[y = 12000e^{-0.20t}\]Step 2: Find the value after 3 years (\(t = 3\)):
\[y = 12000e^{-0.20(3)} = 12000e^{-0.60} = 12000(0.54881...) \approx 6{,}586\]Answer: \(y = 12000e^{-0.20t}\); after 3 years the computer system is worth approximately $6,586.
Verification: \(e^{-0.60} = 0.54881...\), \(12000 \times 0.54881 = 6585.7 \approx 6586\) ✓