Practice Problems - Module 4

Problem 1: Evaluating Exponential Expressions

Evaluate the following expressions without a calculator:

a) 2⁵

b) 3^-2

c) 4^3/2

d) (1/2)^-3

Solution:

Part a) 2⁵
2⁵ = 2 × 2 × 2 × 2 × 2 = 32

Part b) 3^-2
Using the negative exponent rule: a^-n = 1/aⁿ
3^-2 = 1/3² = 1/9

Part c) 4^3/2
Using the fractional exponent rule: a^m/n = (a^1/n)^m
4^3/2 = (4^1/2)³ = 2³ = 8
Or equivalently: 4^3/2 = (√4)³ = 2³ = 8

Part d) (1/2)^-3
Using the negative exponent rule:
(1/2)^-3 = 1/(1/2)³ = 1/(1/8) = 8
Or: (1/2)^-3 = 2³ = 8

Final Answers: a) 32, b) 1/9, c) 8, d) 8

Problem 2: Graphing Exponential Functions

Consider the function f(x) = 2^x.

a) Identify whether this is exponential growth or decay.

b) State the domain and range.

c) Identify the y-intercept.

d) Describe the horizontal asymptote.

e) Find f(0), f(2), and f(-1).

Solution:

Part a) Growth or Decay?
Since the base b = 2 > 1, this is exponential growth.
As x increases, f(x) increases.

Part b) Domain and Range
Domain: All real numbers, (-∞, ∞)
Range: All positive real numbers, (0, ∞)
Note: Exponential functions are never zero or negative.

Part c) Y-intercept
The y-intercept occurs when x = 0:
f(0) = 2⁰ = 1
Y-intercept: (0, 1)

Part d) Horizontal Asymptote
As x → -∞, f(x) → 0
Horizontal asymptote: y = 0 (the x-axis)

Part e) Function Values
f(0) = 2⁰ = 1
f(2) = 2² = 4
f(-1) = 2^-1 = 1/2 = 0.5

Final Answers: a) Exponential growth, b) Domain: (-∞, ∞), Range: (0, ∞), c) (0, 1), d) y = 0, e) f(0) = 1, f(2) = 4, f(-1) = 0.5

Problem 3: Identifying Growth vs Decay

Determine whether each function represents exponential growth or decay, and identify the growth/decay factor:

a) f(x) = 5(1.08)^x

b) g(x) = 100(0.85)^x

c) h(x) = 3(2/3)^x

d) k(x) = 0.5(3)^x

Solution:

Key Rule: For f(x) = a·b^x:
- If b > 1, the function shows exponential growth
- If 0 < b < 1, the function shows exponential decay
The value of a (initial value) doesn't affect growth/decay classification.

Part a) f(x) = 5(1.08)^x
Base b = 1.08 > 1
This is exponential growth.
Growth factor: 1.08 (which represents 8% growth per period)

Part b) g(x) = 100(0.85)^x
Base b = 0.85 < 1
This is exponential decay.
Decay factor: 0.85 (which represents 15% decay per period)

Part c) h(x) = 3(2/3)^x
Base b = 2/3 ≈ 0.667 < 1
This is exponential decay.
Decay factor: 2/3 (which represents 33.3% decay per period)

Part d) k(x) = 0.5(3)^x
Base b = 3 > 1
This is exponential growth.
Growth factor: 3 (which represents 200% growth per period)
Note: Even though a = 0.5 < 1, the base determines growth/decay.

Final Answers: a) Growth, factor 1.08; b) Decay, factor 0.85; c) Decay, factor 2/3; d) Growth, factor 3

Problem 4: Transformations of Exponential Functions

Given the parent function f(x) = 2^x, describe the transformations for each function and state the new horizontal asymptote:

a) g(x) = 2^x+3

b) h(x) = 2^x - 4

c) k(x) = -2^x

d) m(x) = 2^-x + 1

Solution:

Part a) g(x) = 2^x+3
Transformation: Horizontal shift left 3 units
(Adding inside the exponent shifts left)
Horizontal asymptote: y = 0 (unchanged from parent function)

Part b) h(x) = 2^x - 4
Transformation: Vertical shift down 4 units
(Subtracting outside the function shifts down)
Horizontal asymptote: y = -4 (shifted down from y = 0)

Part c) k(x) = -2^x
Transformation: Reflection across the x-axis
(Negative sign in front reflects vertically)
Horizontal asymptote: y = 0 (unchanged)
Note: The function now approaches 0 from below as x → ∞

Part d) m(x) = 2^-x + 1
Transformations:
1) Reflection across the y-axis (negative in exponent)
2) Vertical shift up 1 unit
Note: 2^-x = (1/2)^x, which is exponential decay
Horizontal asymptote: y = 1 (shifted up from y = 0)

Final Answers: a) Left 3, y = 0; b) Down 4, y = -4; c) Reflect over x-axis, y = 0; d) Reflect over y-axis and up 1, y = 1

Problem 5: Compound Interest

You invest $5,000 in an account that pays 6% annual interest compounded quarterly.

a) Write the compound interest formula for this situation.

b) How much money will be in the account after 5 years?

c) How long will it take for the investment to double?

Solution:

Compound Interest Formula: A = P(1 + r/n)^nt
Where: P = principal, r = annual rate (as decimal), n = compounds per year, t = time in years

Part a) Write the Formula
Given: P = $5,000, r = 0.06, n = 4 (quarterly)
A(t) = 5000(1 + 0.06/4)^4t
A(t) = 5000(1.015)^4t

Part b) Amount After 5 Years
Substitute t = 5:
A(5) = 5000(1.015)⁴⁽⁵⁾
A(5) = 5000(1.015)²⁰
A(5) = 5000(1.34685...)
A(5) ≈ $6,734.28

Part c) Time to Double
We want A = $10,000 (double the initial $5,000)
10,000 = 5000(1.015)^4t
Divide both sides by 5000:
2 = (1.015)^4t
Take natural log of both sides:
ln(2) = ln[(1.015)^4t]
ln(2) = 4t · ln(1.015)
t = ln(2) / [4 · ln(1.015)]
t = 0.6931 / [4 × 0.01489]
t = 0.6931 / 0.05956
t ≈ 11.64 years

Final Answers: a) A(t) = 5000(1.015)^4t; b) $6,734.28; c) Approximately 11.64 years

Problem 6: Evaluating Logarithms

Evaluate each logarithm without a calculator:

a) log₂(16)

b) log₃(1/9)

c) log(1000)

d) ln(e³)

e) log₅(1)

Solution:

Remember: log_b(x) = y means b^y = x
Ask yourself: "What power of the base gives me the argument?"

Part a) log₂(16)
Ask: 2 to what power equals 16?
2⁴ = 16
Therefore, log₂(16) = 4

Part b) log₃(1/9)
Ask: 3 to what power equals 1/9?
3^-2 = 1/3² = 1/9
Therefore, log₃(1/9) = -2

Part c) log(1000)
Note: log with no base means log₁₀ (common logarithm)
Ask: 10 to what power equals 1000?
10³ = 1000
Therefore, log(1000) = 3

Part d) ln(e³)
Note: ln means log_e (natural logarithm)
Ask: e to what power equals e³?
The answer is clearly 3
Therefore, ln(e³) = 3

Part e) log₅(1)
Ask: 5 to what power equals 1?
5⁰ = 1 (any non-zero number to the 0 power is 1)
Therefore, log₅(1) = 0
Note: log_b(1) = 0 for any base b

Final Answers: a) 4, b) -2, c) 3, d) 3, e) 0

Problem 7: Expanding Logarithms

Use properties of logarithms to expand each expression completely:

a) log₂(8x)

b) ln(x³/y)

c) log(√(x²y))

d) ln[(x²√y)/z³]

Solution:

Logarithm Properties:
Product Rule: log_b(MN) = log_b(M) + log_b(N)
Quotient Rule: log_b(M/N) = log_b(M) - log_b(N)
Power Rule: log_b(M^p) = p·log_b(M)

Part a) log₂(8x)
Apply product rule:
log₂(8x) = log₂(8) + log₂(x)
Simplify log₂(8): 2³ = 8, so log₂(8) = 3
Final: log₂(8x) = 3 + log₂(x)

Part b) ln(x³/y)
Apply quotient rule:
ln(x³/y) = ln(x³) - ln(y)
Apply power rule to ln(x³):
ln(x³/y) = 3ln(x) - ln(y)

Part c) log(√(x²y))
Rewrite square root as exponent 1/2:
log(√(x²y)) = log[(x²y)^1/2]
Apply power rule:
= (1/2)log(x²y)
Apply product rule:
= (1/2)[log(x²) + log(y)]
Apply power rule to log(x²):
= (1/2)[2log(x) + log(y)]
Distribute:
= log(x) + (1/2)log(y)

Part d) ln[(x²√y)/z³]
Apply quotient rule:
ln[(x²√y)/z³] = ln(x²√y) - ln(z³)
Apply product rule to numerator:
= ln(x²) + ln(√y) - ln(z³)
Rewrite √y = y^1/2 and apply power rule:
= 2ln(x) + (1/2)ln(y) - 3ln(z)

Final Answers: a) 3 + log₂(x); b) 3ln(x) - ln(y); c) log(x) + (1/2)log(y); d) 2ln(x) + (1/2)ln(y) - 3ln(z)

Problem 8: Condensing Logarithms

Use properties of logarithms to condense each expression into a single logarithm:

a) log₃(x) + log₃(y)

b) 2ln(x) - ln(y)

c) 4log(x) + 3log(y) - log(z)

d) (1/2)ln(x) - 2ln(y) + ln(z)

Solution:

Work backwards from expansion:
- Coefficients become exponents (power rule)
- Addition becomes multiplication (product rule)
- Subtraction becomes division (quotient rule)

Part a) log₃(x) + log₃(y)
Apply product rule in reverse:
log₃(x) + log₃(y) = log₃(xy)

Part b) 2ln(x) - ln(y)
Apply power rule to move coefficient:
2ln(x) - ln(y) = ln(x²) - ln(y)
Apply quotient rule:
= ln(x²/y)

Part c) 4log(x) + 3log(y) - log(z)
Apply power rule to all terms:
4log(x) + 3log(y) - log(z) = log(x⁴) + log(y³) - log(z)
Apply product rule to first two terms:
= log(x⁴y³) - log(z)
Apply quotient rule:
= log(x⁴y³/z)

Part d) (1/2)ln(x) - 2ln(y) + ln(z)
Apply power rule to all terms with coefficients:
(1/2)ln(x) - 2ln(y) + ln(z) = ln(x^1/2) - ln(y²) + ln(z)
Apply product rule to ln(x^1/2) and ln(z):
= ln(x^1/2·z) - ln(y²)
Apply quotient rule:
= ln(x^1/2z/y²)
Can also write as: ln(z√x/y²)

Final Answers: a) log₃(xy); b) ln(x²/y); c) log(x⁴y³/z); d) ln(z√x/y²)

Problem 9: Change of Base Formula

Use the change of base formula to evaluate each logarithm. Round to 4 decimal places.

a) log₂(10)

b) log₅(20)

c) log₇(100)

Solution:

Change of Base Formula: log_b(x) = ln(x)/ln(b) or log(x)/log(b)
This allows us to evaluate logarithms with any base using a calculator.

Part a) log₂(10)
Apply change of base formula using natural log:
log₂(10) = ln(10)/ln(2)
= 2.3026/0.6931
≈ 3.3219

Part b) log₅(20)
Apply change of base formula:
log₅(20) = ln(20)/ln(5)
= 2.9957/1.6094
≈ 1.8614

Part c) log₇(100)
Apply change of base formula:
log₇(100) = ln(100)/ln(7)
= 4.6052/1.9459
≈ 2.3674

To verify: 2^3.3219 ≈ 10, 5^1.8614 ≈ 20, 7^2.3674 ≈ 100

Final Answers: a) 3.3219; b) 1.8614; c) 2.3674

Problem 10: Exponential Equations (Same Base)

Solve each equation by writing both sides with the same base:

a) 3^x = 27

b) 2^x+1 = 16

c) 5^2x = 125

d) 4^x = 1/16

Solution:

Strategy: If b^m = bⁿ, then m = n (one-to-one property)
Rewrite both sides with the same base, then equate exponents.

Part a) 3^x = 27
Rewrite 27 as a power of 3: 27 = 3³
3^x = 3³
Equate exponents: x = 3

Part b) 2^x+1 = 16
Rewrite 16 as a power of 2: 16 = 2⁴
2^x+1 = 2⁴
Equate exponents: x + 1 = 4
Solve for x: x = 3

Part c) 5^2x = 125
Rewrite 125 as a power of 5: 125 = 5³
5^2x = 5³
Equate exponents: 2x = 3
Solve for x: x = 3/2 or 1.5

Part d) 4^x = 1/16
Rewrite both as powers of 4:
1/16 = 1/4² = 4^-2
4^x = 4^-2
Equate exponents: x = -2

Final Answers: a) x = 3; b) x = 3; c) x = 3/2; d) x = -2

Problem 11: Exponential Equations (Different Bases)

Solve each equation using logarithms. Round to 4 decimal places where necessary.

a) 3^x = 50

b) 2^x-1 = 10

c) 5 · 2^x = 100

Solution:

Strategy when bases can't match:
1. Isolate the exponential expression
2. Take the logarithm of both sides
3. Use the power rule to bring down the exponent
4. Solve for the variable

Part a) 3^x = 50
Take natural log of both sides:
ln(3^x) = ln(50)
Apply power rule:
x · ln(3) = ln(50)
Divide by ln(3):
x = ln(50)/ln(3)
x = 3.9120/1.0986
x ≈ 3.5609

Part b) 2^x-1 = 10
Take natural log of both sides:
ln(2^x-1) = ln(10)
Apply power rule:
(x - 1) · ln(2) = ln(10)
Divide by ln(2):
x - 1 = ln(10)/ln(2)
x - 1 = 2.3026/0.6931
x - 1 ≈ 3.3219
x ≈ 4.3219

Part c) 5 · 2^x = 100
First, isolate the exponential:
2^x = 100/5
2^x = 20
Take natural log of both sides:
ln(2^x) = ln(20)
Apply power rule:
x · ln(2) = ln(20)
Divide by ln(2):
x = ln(20)/ln(2)
x = 2.9957/0.6931
x ≈ 4.3219

Final Answers: a) x ≈ 3.5609; b) x ≈ 4.3219; c) x ≈ 4.3219

Problem 12: Logarithmic Equations

Solve each logarithmic equation:

a) log₂(x) = 5

b) ln(x) + ln(3) = ln(21)

c) log(x + 2) = 2

d) log₃(2x - 1) = 2

Solution:

Strategy for logarithmic equations:
1. Use log properties to combine terms if possible
2. Convert to exponential form: if log_b(x) = y, then b^y = x
3. Solve the resulting equation
4. Check that solutions make the original logarithms defined

Part a) log₂(x) = 5
Convert to exponential form:
2⁵ = x
x = 32
Check: log₂(32) = log₂(2⁵) = 5

Part b) ln(x) + ln(3) = ln(21)
Use product rule on left side:
ln(3x) = ln(21)
Since the logarithms are equal and have the same base:
3x = 21
x = 7
Check: ln(7) + ln(3) = ln(21)

Part c) log(x + 2) = 2
Convert to exponential form (base 10):
10² = x + 2
100 = x + 2
x = 98
Check: log(98 + 2) = log(100) = 2

Part d) log₃(2x - 1) = 2
Convert to exponential form:
3² = 2x - 1
9 = 2x - 1
10 = 2x
x = 5
Check: log₃(2(5) - 1) = log₃(9) = log₃(3²) = 2

Final Answers: a) x = 32; b) x = 7; c) x = 98; d) x = 5

Problem 13: Checking for Extraneous Solutions

Solve the equation and check for extraneous solutions:

log₂(x) + log₂(x - 3) = 2

Solution:

Step 1: Combine logarithms using product rule
log₂(x) + log₂(x - 3) = 2
log₂[x(x - 3)] = 2
log₂(x² - 3x) = 2

Step 2: Convert to exponential form
2² = x² - 3x
4 = x² - 3x

Step 3: Solve the quadratic equation
x² - 3x - 4 = 0
Factor: (x - 4)(x + 1) = 0
x = 4 or x = -1

Step 4: Check both solutions

Check x = 4:
log₂(4) + log₂(4 - 3) = log₂(4) + log₂(1)
= 2 + 0 = 2
Both arguments (4 and 1) are positive, so x = 4 is valid.

Check x = -1:
log₂(-1) + log₂(-1 - 3) = log₂(-1) + log₂(-4)
These are undefined! Logarithms of negative numbers don't exist in real numbers.
x = -1 is an extraneous solution.

Important: Always check that arguments of logarithms are positive!
For this problem, we need x > 0 AND x - 3 > 0, which means x > 3.

Final Answer: x = 4 (x = -1 is extraneous)

Problem 14: Mixed Exponential and Logarithmic Equation

Solve: e^2x - 5e^x + 6 = 0

Solution:

Strategy: This is a quadratic in form. Let u = e^x, then e^2x = (e^x)² = u²

Step 1: Substitute u = e^x
e^2x - 5e^x + 6 = 0
u² - 5u + 6 = 0

Step 2: Factor the quadratic
u² - 5u + 6 = 0
(u - 2)(u - 3) = 0
u = 2 or u = 3

Step 3: Substitute back e^x = u

Case 1: e^x = 2
Take natural log of both sides:
ln(e^x) = ln(2)
x = ln(2)
x ≈ 0.6931

Case 2: e^x = 3
Take natural log of both sides:
ln(e^x) = ln(3)
x = ln(3)
x ≈ 1.0986

Step 4: Verify solutions
Both solutions are valid since e^x is always positive and both 2 and 3 are positive.

Final Answers: x = ln(2) ≈ 0.6931 or x = ln(3) ≈ 1.0986

Problem 15: Logarithmic Equation with Quotient

Solve: log₅(x + 6) - log₅(x) = 2

Solution:

Step 1: Apply quotient rule
log₅(x + 6) - log₅(x) = 2
log₅[(x + 6)/x] = 2

Step 2: Convert to exponential form
5² = (x + 6)/x
25 = (x + 6)/x

Step 3: Solve for x
Multiply both sides by x:
25x = x + 6
25x - x = 6
24x = 6
x = 6/24 = 1/4 = 0.25

Step 4: Check the solution
We need x > 0 AND x + 6 > 0
Since x = 0.25 > 0, both conditions are satisfied.

Verify: log₅(0.25 + 6) - log₅(0.25)
= log₅(6.25) - log₅(0.25)
= log₅(6.25/0.25)
= log₅(25)
= log₅(5²)
= 2

Final Answer: x = 1/4 or 0.25

Problem 16: Population Growth

A population of bacteria grows according to the formula P(t) = 1000e^0.08t, where t is time in hours.

a) What is the initial population?

b) What is the population after 10 hours?

c) How long will it take for the population to reach 5000?

Solution:

Part a) Initial Population
The initial population occurs at t = 0:
P(0) = 1000e^0.08(0)
P(0) = 1000e⁰
P(0) = 1000(1)
P(0) = 1000 bacteria

Part b) Population After 10 Hours
P(10) = 1000e^0.08(10)
P(10) = 1000e^0.8
P(10) = 1000(2.2255...)
P(10) ≈ 2,226 bacteria

Part c) Time to Reach 5000
Set P(t) = 5000 and solve for t:
5000 = 1000e^0.08t
Divide by 1000:
5 = e^0.08t
Take natural log of both sides:
ln(5) = ln(e^0.08t)
ln(5) = 0.08t
t = ln(5)/0.08
t = 1.6094/0.08
t ≈ 20.12 hours

Final Answers: a) 1000 bacteria; b) Approximately 2,226 bacteria; c) Approximately 20.12 hours

Problem 17: Radioactive Decay

The amount of radioactive substance remaining after t years is given by A(t) = 100e^-0.0125t grams.

a) How much substance remains after 50 years?

b) Find the half-life of the substance (time for half to decay).

Solution:

Part a) Amount After 50 Years
A(50) = 100e^-0.0125(50)
A(50) = 100e^-0.625
A(50) = 100(0.5353...)
A(50) ≈ 53.53 grams

Part b) Half-life
At half-life, A(t) = 50 (half of the original 100 grams)
50 = 100e^-0.0125t
Divide by 100:
0.5 = e^-0.0125t
Take natural log of both sides:
ln(0.5) = ln(e^-0.0125t)
ln(0.5) = -0.0125t
t = ln(0.5)/(-0.0125)
t = -0.6931/(-0.0125)
t ≈ 55.45 years

Alternative formula for half-life: t_1/2 = ln(2)/k
where k is the decay constant (0.0125 in this problem)
t_1/2 = 0.6931/0.0125 ≈ 55.45 years

Final Answers: a) Approximately 53.53 grams; b) Approximately 55.45 years

Problem 18: Compound Interest Application

You want to have $20,000 in 8 years for a down payment on a house. How much should you invest now at 5.5% annual interest compounded:

a) Quarterly?

b) Continuously?

Solution:

Formulas:
Periodic compounding: A = P(1 + r/n)^nt
Continuous compounding: A = Pe^rt
We know A and need to find P (present value).

Part a) Quarterly Compounding
Given: A = $20,000, r = 0.055, n = 4, t = 8
20,000 = P(1 + 0.055/4)⁴⁽⁸⁾
20,000 = P(1.01375)³²
20,000 = P(1.5639...)
P = 20,000/1.5639
P ≈ $12,788.21

You should invest $12,788.21 now.

Part b) Continuous Compounding
Given: A = $20,000, r = 0.055, t = 8
20,000 = Pe^0.055(8)
20,000 = Pe^0.44
20,000 = P(1.5527...)
P = 20,000/1.5527
P ≈ $12,882.50

You should invest $12,882.50 now.

Notice that continuous compounding requires slightly more initial investment because it grows slightly faster, so less is needed initially to reach the same goal.

Final Answers: a) $12,788.21; b) $12,882.50

Problem 19: pH Calculations

The pH of a solution is given by pH = -log[H⁺], where [H⁺] is the hydrogen ion concentration in moles per liter.

a) Find the pH of a solution with [H⁺] = 1.5 × 10^-5

b) Find the hydrogen ion concentration of a solution with pH = 8.3

c) Which solution is more acidic: pH = 3 or pH = 5?

Solution:

pH Scale:
- pH < 7: acidic
- pH = 7: neutral
- pH > 7: basic
Lower pH means more acidic (higher [H⁺])

Part a) Find pH
pH = -log[H⁺]
pH = -log(1.5 × 10^-5)
pH = -[log(1.5) + log(10^-5)]
pH = -[0.1761 + (-5)]
pH = -[0.1761 - 5]
pH = -[-4.8239]
pH ≈ 4.82

Part b) Find [H⁺]
Given: pH = 8.3
8.3 = -log[H⁺]
Multiply by -1:
-8.3 = log[H⁺]
Convert to exponential form (base 10):
[H⁺] = 10^-8.3
[H⁺] ≈ 5.01 × 10^-9 moles per liter

Part c) Which is More Acidic?
pH = 3 is more acidic than pH = 5

Explanation: Lower pH means higher acidity.
The pH = 3 solution has [H⁺] = 10^-3 = 0.001
The pH = 5 solution has [H⁺] = 10^-5 = 0.00001

The pH = 3 solution is 100 times more acidic!

Final Answers: a) pH ≈ 4.82; b) [H⁺] ≈ 5.01 × 10^-9; c) pH = 3 is more acidic

Problem 20: Newton's Law of Cooling

A cup of coffee at 95°C is placed in a room at 20°C. After 10 minutes, the temperature is 70°C. Newton's Law of Cooling states: T(t) = T_room + (T₀ - T_room)e^-kt

a) Find the cooling constant k.

b) What will the temperature be after 20 minutes?

c) How long until the coffee reaches 40°C?

Solution:

Given information:
T_room = 20°C
T₀ = 95°C (initial temperature)
T(10) = 70°C

Part a) Find k
T(t) = 20 + (95 - 20)e^-kt
T(t) = 20 + 75e^-kt

Use the condition T(10) = 70:
70 = 20 + 75e^-10k
50 = 75e^-10k
50/75 = e^-10k
2/3 = e^-10k
Take natural log:
ln(2/3) = -10k
k = -ln(2/3)/10
k = -(-0.4055)/10
k ≈ 0.0405 per minute

Part b) Temperature After 20 Minutes
T(20) = 20 + 75e^-0.0405(20)
T(20) = 20 + 75e^-0.81
T(20) = 20 + 75(0.4449)
T(20) = 20 + 33.37
T(20) ≈ 53.4°C

Part c) Time to Reach 40°C
40 = 20 + 75e^-0.0405t
20 = 75e^-0.0405t
20/75 = e^-0.0405t
4/15 = e^-0.0405t
Take natural log:
ln(4/15) = -0.0405t
t = ln(4/15)/(-0.0405)
t = -1.3218/(-0.0405)
t ≈ 32.64 minutes

Final Answers: a) k ≈ 0.0405 per minute; b) Approximately 53.4°C; c) Approximately 32.64 minutes

Module 4: Practice Problems

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