Lesson 4: Applications of Exponential and Logarithmic Functions
Introduction
Exponential and logarithmic functions are not just abstract mathematical concepts—they model many real-world phenomena that surround us every day. From the growth of populations and investments to the decay of radioactive substances and the cooling of hot coffee, these functions provide powerful tools for understanding and predicting natural and man-made processes.
In this lesson, we will explore various applications of exponential and logarithmic functions across multiple fields including biology, finance, physics, and chemistry. You will learn to identify which situations involve exponential growth versus decay, select appropriate formulas, interpret parameters, and solve practical problems using the techniques developed in previous lessons.
Learning Objectives:
- Model population growth using exponential functions
- Solve radioactive decay and half-life problems
- Apply compound interest formulas to financial situations
- Use Newton's Law of Cooling for temperature problems
- Work with logarithmic scales in chemistry and physics
- Interpret and solve real-world problems involving exponential and logarithmic relationships
Section 1: Population Growth
Population growth often follows an exponential pattern when resources are abundant and growth is unrestricted. Whether modeling bacteria in a petri dish, human populations in cities, or animal populations in ecosystems, exponential growth models provide valuable insights.
Population Growth Models
Continuous Growth Model: P(t) = P₀ · e^(rt)
Discrete Growth Model: P(t) = P₀ · (1 + r)^t
Where:
- P(t) = population at time t
- P₀ = initial population (at t = 0)
- r = growth rate (as a decimal)
- t = time
- e ≈ 2.71828
Doubling Time Formula: t_double = ln(2)/r (for continuous growth)
Example 1: Finding Growth Rate from Data
Problem: A city's population grows from 50,000 to 75,000 in 10 years. Assuming continuous exponential growth, find the annual growth rate.
Given:
- P₀ = 50,000
- P(10) = 75,000
- t = 10 years
Step 1: Use the continuous growth model P(t) = P₀ · e^(rt)
75,000 = 50,000 · e^(10r)
Step 2: Divide both sides by 50,000
1.5 = e^(10r)
Step 3: Take the natural logarithm of both sides
ln(1.5) = ln(e^(10r))
ln(1.5) = 10r
Step 4: Solve for r
r = ln(1.5)/10
r ≈ 0.4055/10
r ≈ 0.04055
Answer: The annual growth rate is approximately 4.055% or 4.06%
Example 2: Predicting Future Population
Problem: Using the growth rate from Example 1, predict the city's population in 20 years from the initial time.
Given:
- P₀ = 50,000
- r ≈ 0.04055
- t = 20 years
Step 1: Use P(t) = P₀ · e^(rt)
P(20) = 50,000 · e^(0.04055 × 20)
Step 2: Calculate the exponent
P(20) = 50,000 · e^0.811
Step 3: Evaluate e^0.811
P(20) = 50,000 · 2.250
Step 4: Multiply
P(20) = 112,500
Answer: The predicted population in 20 years is approximately 112,500 people
Example 3: Finding When Population Reaches a Target
Problem: Using the same growth rate (r ≈ 0.04055), when will the city's population reach 100,000?
Given:
- P₀ = 50,000
- P(t) = 100,000
- r ≈ 0.04055
Step 1: Set up the equation
100,000 = 50,000 · e^(0.04055t)
Step 2: Divide by 50,000
2 = e^(0.04055t)
Step 3: Take natural logarithm of both sides
ln(2) = 0.04055t
Step 4: Solve for t
t = ln(2)/0.04055
t ≈ 0.6931/0.04055
t ≈ 17.09 years
Answer: The population will reach 100,000 in approximately 17.1 years
Note: This is the doubling time, which makes sense since 100,000 is double the initial 50,000
Example 4: Bacteria Growth with Doubling Time
Problem: A bacteria culture doubles every 3 hours. If there are initially 500 bacteria, how many will there be after 12 hours?
Method 1: Using doubling directly
In 12 hours, there are 12/3 = 4 doubling periods
After each doubling: 500 → 1,000 → 2,000 → 4,000 → 8,000
Or use: P(12) = 500 · 2^4 = 500 · 16 = 8,000 bacteria
Method 2: Finding growth rate first
Step 1: Use doubling time formula to find r
t_double = ln(2)/r
3 = ln(2)/r
r = ln(2)/3 ≈ 0.2310 per hour
Step 2: Use P(t) = P₀ · e^(rt)
P(12) = 500 · e^(0.2310 × 12)
P(12) = 500 · e^2.772
P(12) = 500 · 16
P(12) = 8,000 bacteria
Answer: There will be 8,000 bacteria after 12 hours
Example 5: Comparing Growth Rates
Problem: Town A has a population of 10,000 growing at 2% per year. Town B has a population of 15,000 growing at 1.5% per year. When will Town A's population equal Town B's population?
Given:
- Town A: P_A(t) = 10,000 · e^(0.02t)
- Town B: P_B(t) = 15,000 · e^(0.015t)
Step 1: Set populations equal
10,000 · e^(0.02t) = 15,000 · e^(0.015t)
Step 2: Divide both sides by 10,000
e^(0.02t) = 1.5 · e^(0.015t)
Step 3: Divide both sides by e^(0.015t)
e^(0.02t)/e^(0.015t) = 1.5
e^(0.02t - 0.015t) = 1.5
e^(0.005t) = 1.5
Step 4: Take natural logarithm
0.005t = ln(1.5)
0.005t ≈ 0.4055
Step 5: Solve for t
t ≈ 0.4055/0.005
t ≈ 81.1 years
Answer: Town A will catch up to Town B in approximately 81.1 years
Verification: P_A(81.1) ≈ 10,000 · e^1.622 ≈ 50,700 and P_B(81.1) ≈ 15,000 · e^1.217 ≈ 50,700
Section 2: Radioactive Decay and Half-Life
Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This process follows an exponential decay pattern, which is the opposite of exponential growth. The concept of half-life—the time required for half of the substance to decay—is fundamental in nuclear physics, archaeology, and medicine.
Radioactive Decay Models
Continuous Decay Model: A(t) = A₀ · e^(-kt)
Half-Life Model: A(t) = A₀ · (1/2)^(t/h)
Where:
- A(t) = amount remaining at time t
- A₀ = initial amount
- k = decay constant (positive number)
- h = half-life
- t = time
Relationship: h = ln(2)/k or k = ln(2)/h
Example 6: Finding Decay Rate from Data
Problem: A 100-gram sample of a radioactive substance decays to 75 grams in 5 years. Find the decay constant k.
Given:
- A₀ = 100 grams
- A(5) = 75 grams
- t = 5 years
Step 1: Use A(t) = A₀ · e^(-kt)
75 = 100 · e^(-5k)
Step 2: Divide by 100
0.75 = e^(-5k)
Step 3: Take natural logarithm
ln(0.75) = -5k
Step 4: Solve for k
k = -ln(0.75)/5
k = -(-0.2877)/5
k ≈ 0.0575 per year
Answer: The decay constant is approximately 0.0575 per year, or 5.75% per year
Example 7: Predicting Amount Remaining
Problem: Using the decay constant from Example 6, how much of the original 100 grams will remain after 20 years?
Given:
- A₀ = 100 grams
- k ≈ 0.0575
- t = 20 years
Step 1: Use A(t) = A₀ · e^(-kt)
A(20) = 100 · e^(-0.0575 × 20)
Step 2: Calculate the exponent
A(20) = 100 · e^(-1.15)
Step 3: Evaluate
A(20) = 100 · 0.3166
A(20) ≈ 31.66 grams
Answer: Approximately 31.66 grams will remain after 20 years
Example 8: Finding Half-Life from Decay Rate
Problem: If a radioactive substance has a decay constant k = 0.0575 per year, what is its half-life?
Given: k = 0.0575 per year
Step 1: Use half-life formula h = ln(2)/k
h = ln(2)/0.0575
Step 2: Calculate
h = 0.6931/0.0575
h ≈ 12.05 years
Answer: The half-life is approximately 12.05 years
Verification: After 12.05 years, A(12.05) = 100 · e^(-0.0575 × 12.05) ≈ 50 grams
Example 9: Carbon-14 Dating
Problem: Carbon-14 has a half-life of 5,730 years. An ancient wooden artifact contains 30% of the original Carbon-14. Estimate the age of the artifact.
Given:
- Half-life h = 5,730 years
- A(t)/A₀ = 0.30 (30% remains)
Method 1: Using half-life model
Step 1: Use A(t) = A₀ · (1/2)^(t/h)
0.30A₀ = A₀ · (1/2)^(t/5730)
Step 2: Divide by A₀
0.30 = (1/2)^(t/5730)
Step 3: Take logarithm of both sides
log(0.30) = (t/5730) · log(1/2)
-0.5229 = (t/5730) · (-0.3010)
Step 4: Solve for t
t = 5730 · 0.5229/0.3010
t ≈ 5730 · 1.737
t ≈ 9,953 years
Method 2: Using decay constant
k = ln(2)/5730 ≈ 0.000121
0.30 = e^(-0.000121t)
ln(0.30) = -0.000121t
t = -ln(0.30)/0.000121 ≈ 9,953 years
Answer: The artifact is approximately 9,953 years old, or about 10,000 years
Example 10: Medical Isotope Decay
Problem: Iodine-131, used in medical treatments, has a half-life of 8 days. A hospital receives a 20-milligram sample. How much will remain after 30 days? Is this enough for treatment if 2 mg is required?
Given:
- A₀ = 20 mg
- h = 8 days
- t = 30 days
Step 1: Use A(t) = A₀ · (1/2)^(t/h)
A(30) = 20 · (1/2)^(30/8)
Step 2: Calculate the exponent
A(30) = 20 · (1/2)^3.75
Step 3: Evaluate (1/2)^3.75
(1/2)^3.75 = 2^(-3.75) ≈ 0.0745
Step 4: Multiply
A(30) = 20 · 0.0745
A(30) ≈ 1.49 mg
Answer: After 30 days, approximately 1.49 mg remains. This is NOT enough for treatment since 2 mg is required. The sample should be used within about 27 days to have at least 2 mg available.
Section 3: Compound Interest and Finance
One of the most practical applications of exponential functions is in finance, particularly in calculating compound interest. Understanding how money grows over time with different compounding frequencies is essential for making informed financial decisions about savings, investments, and loans.
Compound Interest Formulas
Periodic Compounding: A = P(1 + r/n)^(nt)
Continuous Compounding: A = Pe^(rt)
Where:
- A = final amount
- P = principal (initial investment)
- r = annual interest rate (as decimal)
- n = number of times compounded per year
- t = time in years
Effective Annual Rate: EAR = (1 + r/n)^n - 1
Common Compounding Frequencies:
- Annually: n = 1
- Semi-annually: n = 2
- Quarterly: n = 4
- Monthly: n = 12
- Daily: n = 365
- Continuously: use A = Pe^(rt)
Example 11: Quarterly Compounding
Problem: You invest $5,000 in an account that pays 6% annual interest compounded quarterly. How much will you have after 10 years?
Given:
- P = $5,000
- r = 0.06
- n = 4 (quarterly)
- t = 10 years
Step 1: Use A = P(1 + r/n)^(nt)
A = 5,000(1 + 0.06/4)^(4 × 10)
Step 2: Simplify inside parentheses
A = 5,000(1 + 0.015)^40
A = 5,000(1.015)^40
Step 3: Calculate (1.015)^40
(1.015)^40 ≈ 1.8140
Step 4: Multiply by principal
A = 5,000 · 1.8140
A ≈ $9,070.09
Answer: After 10 years, you will have approximately $9,070.09
Interest Earned: $9,070.09 - $5,000 = $4,070.09
Example 12: Comparing Quarterly vs. Continuous Compounding
Problem: Compare the final amounts from Example 11 (quarterly compounding) with continuous compounding at the same rate.
Given: P = $5,000, r = 0.06, t = 10
Quarterly (from Example 11): A ≈ $9,070.09
Continuous Compounding:
Step 1: Use A = Pe^(rt)
A = 5,000 · e^(0.06 × 10)
Step 2: Calculate exponent
A = 5,000 · e^0.6
Step 3: Evaluate e^0.6
A = 5,000 · 1.8221
A ≈ $9,110.59
Comparison:
| Method | Final Amount | Interest Earned |
|---|---|---|
| Quarterly | $9,070.09 | $4,070.09 |
| Continuous | $9,110.59 | $4,110.59 |
| Difference | $40.50 | $40.50 |
Answer: Continuous compounding yields $40.50 more than quarterly compounding over 10 years
Example 13: Finding Time to Double
Problem: How long will it take for an investment to double if it earns 8% annual interest compounded monthly?
Given:
- r = 0.08
- n = 12 (monthly)
- A = 2P (double the principal)
Step 1: Set up equation A = P(1 + r/n)^(nt)
2P = P(1 + 0.08/12)^(12t)
Step 2: Divide by P
2 = (1 + 0.08/12)^(12t)
2 = (1.00667)^(12t)
Step 3: Take logarithm of both sides
log(2) = 12t · log(1.00667)
0.3010 = 12t · 0.002894
Step 4: Solve for t
t = 0.3010/(12 · 0.002894)
t = 0.3010/0.03473
t ≈ 8.67 years
Answer: It will take approximately 8.67 years (or about 8 years 8 months) for the investment to double
Rule of 72 Approximation: 72/8 = 9 years (close estimate)
Example 14: Finding Required Interest Rate
Problem: You want to grow $5,000 to $10,000 in 5 years. What annual interest rate do you need if interest is compounded monthly?
Given:
- P = $5,000
- A = $10,000
- t = 5 years
- n = 12 (monthly)
Step 1: Set up equation
10,000 = 5,000(1 + r/12)^(12 × 5)
Step 2: Divide by 5,000
2 = (1 + r/12)^60
Step 3: Take the 60th root
2^(1/60) = 1 + r/12
1.01162 = 1 + r/12
Step 4: Solve for r
r/12 = 1.01162 - 1
r/12 = 0.01162
r = 0.01162 · 12
r ≈ 0.1394
Answer: You need an annual interest rate of approximately 13.94% compounded monthly
Example 15: Comparing Investment Options Using EAR
Problem: Which is a better investment: Option A offering 6.5% compounded quarterly, or Option B offering 6.4% compounded continuously?
Option A: 6.5% compounded quarterly
Step 1: Calculate EAR using EAR = (1 + r/n)^n - 1
EAR_A = (1 + 0.065/4)^4 - 1
EAR_A = (1.01625)^4 - 1
EAR_A = 1.06660 - 1
EAR_A ≈ 0.06660 or 6.660%
Option B: 6.4% compounded continuously
Step 2: For continuous compounding, EAR = e^r - 1
EAR_B = e^0.064 - 1
EAR_B = 1.06612 - 1
EAR_B ≈ 0.06612 or 6.612%
Comparison:
| Option | Nominal Rate | Compounding | Effective Annual Rate |
|---|---|---|---|
| A | 6.5% | Quarterly | 6.660% |
| B | 6.4% | Continuous | 6.612% |
Answer: Option A is better, yielding an effective annual rate of 6.660% compared to Option B's 6.612%
Note: Even though Option A has a higher nominal rate, this comparison shows the importance of calculating the effective rate to make fair comparisons
Section 4: Newton's Law of Cooling
Newton's Law of Cooling describes how the temperature of an object changes as it approaches the temperature of its surroundings. This exponential model applies to many real-world situations, from cooling coffee to forensic science investigations.
Newton's Law of Cooling
Temperature Model: T(t) = T_a + (T₀ - T_a) · e^(-kt)
Where:
- T(t) = temperature at time t
- T_a = ambient (surrounding) temperature
- T₀ = initial temperature
- k = cooling constant (positive number)
- t = time
Note: The formula works for both cooling (T₀ > T_a) and heating (T₀ < T_a)
Example 16: Coffee Cooling
Problem: A cup of coffee at 90 degrees C is placed in a room at 20 degrees C. After 10 minutes, the coffee has cooled to 70 degrees C. What will the temperature be after 20 minutes?
Given:
- T₀ = 90°C
- T_a = 20°C
- T(10) = 70°C
Step 1: Find the cooling constant k using the 10-minute data
T(t) = T_a + (T₀ - T_a) · e^(-kt)
70 = 20 + (90 - 20) · e^(-10k)
70 = 20 + 70 · e^(-10k)
Step 2: Solve for k
50 = 70 · e^(-10k)
50/70 = e^(-10k)
0.7143 = e^(-10k)
ln(0.7143) = -10k
-0.3365 = -10k
k ≈ 0.03365
Step 3: Find temperature at t = 20
T(20) = 20 + 70 · e^(-0.03365 × 20)
T(20) = 20 + 70 · e^(-0.673)
T(20) = 20 + 70 · 0.5102
T(20) = 20 + 35.71
T(20) ≈ 55.71°C
Answer: After 20 minutes, the coffee will be approximately 55.71°C
Example 17: Finding Cooling Constant from Data
Problem: A baked potato at 200°F is placed in a 70°F room. After 30 minutes, it has cooled to 150°F. Find the cooling constant k.
Given:
- T₀ = 200°F
- T_a = 70°F
- T(30) = 150°F
Step 1: Set up equation
150 = 70 + (200 - 70) · e^(-30k)
150 = 70 + 130 · e^(-30k)
Step 2: Isolate exponential term
80 = 130 · e^(-30k)
80/130 = e^(-30k)
0.6154 = e^(-30k)
Step 3: Take natural logarithm
ln(0.6154) = -30k
-0.4855 = -30k
Step 4: Solve for k
k = 0.4855/30
k ≈ 0.01618 per minute
Answer: The cooling constant is approximately 0.01618 per minute
Example 18: When Will Temperature Reach Target?
Problem: Using the coffee data from Example 16, when will the coffee reach a drinkable temperature of 60°C?
Given:
- T_a = 20°C
- T₀ = 90°C
- k ≈ 0.03365
- T(t) = 60°C
Step 1: Set up equation
60 = 20 + (90 - 20) · e^(-0.03365t)
60 = 20 + 70 · e^(-0.03365t)
Step 2: Isolate exponential
40 = 70 · e^(-0.03365t)
40/70 = e^(-0.03365t)
0.5714 = e^(-0.03365t)
Step 3: Take natural logarithm
ln(0.5714) = -0.03365t
-0.5596 = -0.03365t
Step 4: Solve for t
t = 0.5596/0.03365
t ≈ 16.63 minutes
Answer: The coffee will reach 60°C in approximately 16.6 minutes (about 16 minutes 38 seconds)
Section 5: pH and Chemistry Applications
The pH scale is a logarithmic measure of acidity and alkalinity used extensively in chemistry, biology, and environmental science. Understanding pH calculations requires working with logarithms and recognizing that the scale is logarithmic, meaning each unit change represents a tenfold difference in hydrogen ion concentration.
pH Formulas
pH Definition: pH = -log[H⁺]
Hydrogen Ion Concentration: [H⁺] = 10^(-pH)
Where:
- [H⁺] = hydrogen ion concentration in moles per liter
- log = common logarithm (base 10)
pH Scale:
- pH < 7: Acidic
- pH = 7: Neutral
- pH > 7: Basic (Alkaline)
- Scale typically ranges from 0 to 14
Important: Each pH unit represents a 10-fold change in [H⁺]
Example 19: Finding pH from Hydrogen Ion Concentration
Problem: Find the pH of a solution with hydrogen ion concentration [H⁺] = 1 × 10⁻⁵ M.
Given: [H⁺] = 1 × 10⁻⁵ M
Step 1: Use pH = -log[H⁺]
pH = -log(1 × 10⁻⁵)
Step 2: Apply logarithm properties
pH = -log(10⁻⁵)
pH = -(-5)
pH = 5
Answer: The pH is 5
Classification: This is acidic (pH < 7)
Example 20: Finding Hydrogen Ion Concentration from pH
Problem: A solution has a pH of 3.5. Find the hydrogen ion concentration [H⁺].
Given: pH = 3.5
Step 1: Use [H⁺] = 10^(-pH)
[H⁺] = 10^(-3.5)
Step 2: Calculate
[H⁺] = 10^(-3.5)
[H⁺] ≈ 3.16 × 10⁻⁴ M
Answer: The hydrogen ion concentration is approximately 3.16 × 10⁻⁴ M
Note: This can also be written as 0.000316 M
Example 21: Comparing Acidity Levels
Problem: How many times more acidic is a solution with pH 2 compared to a solution with pH 5?
Given: pH₁ = 2, pH₂ = 5
Step 1: Find [H⁺] for each solution
[H⁺]₁ = 10^(-2) = 0.01 M
[H⁺]₂ = 10^(-5) = 0.00001 M
Step 2: Calculate ratio
Ratio = [H⁺]₁ / [H⁺]₂
Ratio = 0.01 / 0.00001
Ratio = 1,000
Alternative Method:
Difference in pH = 5 - 2 = 3 units
Ratio = 10³ = 1,000
Answer: The pH 2 solution is 1,000 times more acidic than the pH 5 solution
Key Concept: Each pH unit difference represents a factor of 10 in acidity
Example 22: Real-World pH Comparison
Problem: Vinegar has a pH of 2.4 and lemon juice has a pH of 2.0. Which is more acidic and by how much?
Given: pH_vinegar = 2.4, pH_lemon = 2.0
Step 1: Determine which is more acidic
Lower pH means more acidic
2.0 < 2.4, so lemon juice is more acidic
Step 2: Find [H⁺] for each
[H⁺]_lemon = 10^(-2.0) = 0.01 M
[H⁺]_vinegar = 10^(-2.4) ≈ 3.98 × 10⁻³ M
Step 3: Calculate ratio
Ratio = [H⁺]_lemon / [H⁺]_vinegar
Ratio = 0.01 / (3.98 × 10⁻³)
Ratio ≈ 2.51
Alternative Calculation:
Difference = 2.4 - 2.0 = 0.4 pH units
Ratio = 10^0.4 ≈ 2.51
Answer: Lemon juice is more acidic than vinegar by a factor of approximately 2.51, meaning it has about 2.5 times more hydrogen ions
Section 6: Other Applications
Logarithmic and exponential functions appear in many other scientific and real-world contexts. We will explore earthquake magnitude measurement, sound intensity, and logistic growth models that account for limiting factors.
Additional Logarithmic Scales
Richter Scale (Earthquake Magnitude): M = log(I/I₀)
- M = magnitude
- I = intensity of earthquake
- I₀ = reference intensity
- Each unit increase represents 10× intensity
Decibel Scale (Sound Intensity): dB = 10 log(I/I₀)
- dB = decibels
- I = sound intensity
- I₀ = threshold of hearing
- 10 dB increase represents 10× intensity
Logistic Growth Model: P(t) = K / (1 + Ae^(-rt))
- P(t) = population at time t
- K = carrying capacity
- A = constant related to initial population
- r = growth rate
Example 23: Earthquake Intensity Comparison
Problem: An earthquake measures 7.2 on the Richter scale. Another earthquake measures 5.2. How many times more intense is the first earthquake?
Given: M₁ = 7.2, M₂ = 5.2
Step 1: Find intensity for each using M = log(I/I₀)
7.2 = log(I₁/I₀), so I₁/I₀ = 10^7.2
5.2 = log(I₂/I₀), so I₂/I₀ = 10^5.2
Step 2: Calculate ratio
Ratio = I₁/I₂ = (10^7.2)/(10^5.2)
Ratio = 10^(7.2-5.2)
Ratio = 10^2
Ratio = 100
Answer: The 7.2 magnitude earthquake is 100 times more intense than the 5.2 magnitude earthquake
Key Concept: The difference is 2 magnitude units, and 10² = 100
Example 24: Sound Level Calculations
Problem: A normal conversation measures 60 dB. A rock concert measures 110 dB. How many times more intense is the rock concert?
Given: dB₁ = 60, dB₂ = 110
Step 1: Find intensity ratio using dB = 10 log(I/I₀)
60 = 10 log(I₁/I₀), so log(I₁/I₀) = 6
110 = 10 log(I₂/I₀), so log(I₂/I₀) = 11
Step 2: Express intensities
I₁/I₀ = 10⁶
I₂/I₀ = 10¹¹
Step 3: Calculate ratio
Ratio = I₂/I₁ = (10¹¹)/(10⁶)
Ratio = 10⁵
Ratio = 100,000
Answer: The rock concert is 100,000 times more intense than normal conversation
Note: This is why prolonged exposure to loud music can damage hearing
Example 25: Logistic Population Model
Problem: A fish population in a lake follows the logistic model P(t) = 5000 / (1 + 49e^(-0.4t)), where t is in years. Find the initial population and the carrying capacity. What is the population after 5 years?
Given: P(t) = 5000 / (1 + 49e^(-0.4t))
Step 1: Find initial population (t = 0)
P(0) = 5000 / (1 + 49e^0)
P(0) = 5000 / (1 + 49)
P(0) = 5000 / 50
P(0) = 100 fish
Step 2: Identify carrying capacity
As t → ∞, e^(-0.4t) → 0
K = 5000 / (1 + 0) = 5000 fish
Step 3: Find population at t = 5
P(5) = 5000 / (1 + 49e^(-0.4×5))
P(5) = 5000 / (1 + 49e^(-2))
P(5) = 5000 / (1 + 49 × 0.1353)
P(5) = 5000 / (1 + 6.630)
P(5) = 5000 / 7.630
P(5) ≈ 655 fish
Answer:
- Initial population: 100 fish
- Carrying capacity: 5,000 fish
- Population after 5 years: approximately 655 fish
Example 26: Drug Concentration Decay
Problem: After taking medication, the concentration C(t) of a drug in the bloodstream (in mg/L) follows C(t) = 8e^(-0.3t), where t is in hours. What is the initial concentration? When will the concentration drop to 2 mg/L?
Given: C(t) = 8e^(-0.3t)
Step 1: Find initial concentration (t = 0)
C(0) = 8e^0 = 8 × 1 = 8 mg/L
Step 2: Find when C(t) = 2
2 = 8e^(-0.3t)
Step 3: Solve for t
2/8 = e^(-0.3t)
0.25 = e^(-0.3t)
Step 4: Take natural logarithm
ln(0.25) = -0.3t
-1.386 = -0.3t
Step 5: Solve for t
t = 1.386/0.3
t ≈ 4.62 hours
Answer:
- Initial concentration: 8 mg/L
- Time to reach 2 mg/L: approximately 4.62 hours (about 4 hours 37 minutes)
Note: The half-life is ln(2)/0.3 ≈ 2.31 hours, and after two half-lives (4.62 hours), the concentration is 8 × (1/2)² = 2 mg/L
Check Your Understanding
Question 1: A population of 2,000 grows to 3,500 in 8 years. What is the annual growth rate assuming continuous growth?
Use P(t) = P₀e^(rt)
3,500 = 2,000e^(8r)
1.75 = e^(8r)
ln(1.75) = 8r
r = ln(1.75)/8 ≈ 0.0698 or about 6.98%
Question 2: A radioactive isotope has a half-life of 12 days. What percentage of the original amount remains after 30 days?
Use A(t) = A₀(1/2)^(t/h)
A(30) = A₀(1/2)^(30/12)
A(30) = A₀(1/2)^2.5
A(30) = A₀(0.1768)
About 17.68% remains
Question 3: You invest $8,000 at 5.5% compounded monthly. How much will you have after 7 years?
Use A = P(1 + r/n)^(nt)
A = 8,000(1 + 0.055/12)^(12×7)
A = 8,000(1.004583)^84
A = 8,000(1.4743)
A ≈ $11,794.40
Question 4: Which earns more: $10,000 at 6% compounded quarterly for 5 years, or $10,000 at 5.9% compounded continuously for 5 years?
Quarterly: A = 10,000(1.015)^20 ≈ $13,468.55
Continuous: A = 10,000e^(0.059×5) = 10,000e^0.295 ≈ $13,432.50
Quarterly compounding earns more by about $36.05
Question 5: A cup of tea at 95°C is placed in a 22°C room. After 8 minutes it has cooled to 75°C. What will the temperature be after 15 minutes?
First find k: 75 = 22 + (95-22)e^(-8k)
53 = 73e^(-8k), so k ≈ 0.0403
Then: T(15) = 22 + 73e^(-0.0403×15)
T(15) = 22 + 73e^(-0.605) ≈ 62.3°C
Question 6: Find the pH of a solution with [H⁺] = 4.2 × 10⁻⁴ M.
pH = -log[H⁺]
pH = -log(4.2 × 10⁻⁴)
pH = -log(4.2) - log(10⁻⁴)
pH = -0.623 + 4
pH ≈ 3.38 (acidic)
Question 7: An earthquake measures 6.8 on the Richter scale. How much more intense is this than an earthquake measuring 4.8?
Difference = 6.8 - 4.8 = 2 magnitude units
Intensity ratio = 10² = 100
The 6.8 earthquake is 100 times more intense
Question 8: A town's population follows P(t) = 12,000/(1 + 15e^(-0.5t)). What is the carrying capacity? What is the population after 10 years?
Carrying capacity K = 12,000 (numerator)
P(10) = 12,000/(1 + 15e^(-5))
P(10) = 12,000/(1 + 15(0.0067))
P(10) = 12,000/(1.101) ≈ 10,900