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Exponential Functions

Properties and graphs of exponential functions

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Exponential Functions

Definition

An exponential function has the form: f(x)=a⋅bxf(x) = a \cdot b^x

where:

  • aa = initial value (y-intercept when x=0x = 0)
  • bb = base (growth/decay factor)
  • xx = exponent (input variable)

Growth vs. Decay

Exponential Growth: b>1b > 1

  • Function increases
  • Example: f(x)=2xf(x) = 2^x

Exponential Decay: 0<b<10 < b < 1

  • Function decreases
  • Example: f(x)=(12)xf(x) = (\frac{1}{2})^x

Properties

  • Domain: All real numbers
  • Range: (0,∞)(0, \infty) if a>0a > 0
  • Y-intercept: (0,a)(0, a)
  • Horizontal asymptote: y=0y = 0
  • Never touches or crosses x-axis

Exponential Growth/Decay Formula

A=A0(1+r)tA = A_0(1 + r)^t

where:

  • AA = final amount
  • A0A_0 = initial amount
  • rr = rate (as decimal)
  • tt = time

Growth: r>0r > 0 (add) Decay: r<0r < 0 (subtract)

📚 Practice Problems

1Problem 1easy

❓ Question:

Evaluate: 2⁵

💡 Show Solution

Step 1: Calculate the power: 2⁵ = 2 × 2 × 2 × 2 × 2

Step 2: Multiply step by step: 2 × 2 = 4 4 × 2 = 8 8 × 2 = 16 16 × 2 = 32

Answer: 32

2Problem 2easy

❓ Question:

Evaluate: f(x)=3⋅2xf(x) = 3 \cdot 2^x when x=4x = 4

💡 Show Solution

Substitute x=4x = 4 into the function:

f(4)=3⋅24f(4) = 3 \cdot 2^4 =3⋅16= 3 \cdot 16 =48= 48

Answer: f(4)=48f(4) = 48

3Problem 3easy

❓ Question:

Simplify: (3²)(3⁴)

💡 Show Solution

Step 1: Use the product rule for exponents: aᵐ · aⁿ = aᵐ⁺ⁿ

Step 2: Add the exponents: (3²)(3⁴) = 3²⁺⁴ = 3⁶

Step 3: Evaluate (optional): 3⁶ = 729

Answer: 3⁶ or 729

4Problem 4easy

❓ Question:

Evaluate: f(x)=3⋅2xf(x) = 3 \cdot 2^x when x=4x = 4

💡 Show Solution

Substitute x=4x = 4 into the function:

f(4)=3⋅24f(4) = 3 \cdot 2^4 =3⋅16= 3 \cdot 16 =48= 48

Answer: f(4)=48f(4) = 48

5Problem 5medium

❓ Question:

A population of bacteria doubles every 3 hours. If there are initially 500 bacteria, how many will there be after 12 hours?

💡 Show Solution

Step 1: Determine how many doubling periods 12 hours3 hours/doubling=4 doublings\frac{12 \text{ hours}}{3 \text{ hours/doubling}} = 4 \text{ doublings}

Step 2: Use the formula A=A0⋅2nA = A_0 \cdot 2^n A=500⋅24A = 500 \cdot 2^4 =500⋅16= 500 \cdot 16 =8000= 8000

Answer: 8,000 bacteria

6Problem 6medium

❓ Question:

A population of bacteria doubles every 3 hours. If there are initially 500 bacteria, how many will there be after 12 hours?

💡 Show Solution

Step 1: Determine how many doubling periods 12 hours3 hours/doubling=4 doublings\frac{12 \text{ hours}}{3 \text{ hours/doubling}} = 4 \text{ doublings}

Step 2: Use the formula A=A0⋅2nA = A_0 \cdot 2^n A=500⋅24A = 500 \cdot 2^4 =500⋅16= 500 \cdot 16 =8000= 8000

Answer: 8,000 bacteria

7Problem 7medium

❓ Question:

If f(x) = 2ˣ, find f(3), f(-2), and f(0).

💡 Show Solution

Step 1: Find f(3): f(3) = 2³ = 8

Step 2: Find f(-2): f(-2) = 2⁻² = 1/(2²) = 1/4

Step 3: Find f(0): f(0) = 2⁰ = 1

Step 4: Note the pattern:

  • Positive exponent: regular multiplication
  • Negative exponent: reciprocal
  • Zero exponent: always equals 1

Answer: f(3) = 8, f(-2) = 1/4, f(0) = 1

8Problem 8hard

❓ Question:

A car depreciates at 15% per year. If it costs $25,000 new, what will it be worth after 5 years?

💡 Show Solution

Use the decay formula: A=A0(1−r)tA = A_0(1 - r)^t

Given:

  • A0=25000A_0 = 25000
  • r=0.15r = 0.15 (15% decay)
  • t=5t = 5 years

Substitute: A=25000(1−0.15)5A = 25000(1 - 0.15)^5 =25000(0.85)5= 25000(0.85)^5 =25000(0.4437...)= 25000(0.4437...) ≈11,093\approx 11,093

Answer: Approximately $11,093

9Problem 9hard

❓ Question:

A car depreciates at 15% per year. If it costs $25,000 new, what will it be worth after 5 years?

💡 Show Solution

Use the decay formula: A=A0(1−r)tA = A_0(1 - r)^t

Given:

  • A0=25000A_0 = 25000
  • r=0.15r = 0.15 (15% decay)
  • t=5t = 5 years

Substitute: A=25000(1−0.15)5A = 25000(1 - 0.15)^5 =25000(0.85)5= 25000(0.85)^5 =25000(0.4437...)= 25000(0.4437...) ≈11,093\approx 11,093

Answer: Approximately $11,093

10Problem 10medium

❓ Question:

A bacteria population doubles every 3 hours. If there are initially 500 bacteria, write an exponential function P(t) for the population after t hours.

💡 Show Solution

Step 1: Identify the exponential growth formula: P(t) = P₀ · aᵗ/ᵏ

Where:

  • P₀ = initial population
  • a = growth factor
  • k = time period for one growth cycle

Step 2: Identify the values: P₀ = 500 (initial population) a = 2 (doubles) k = 3 (every 3 hours)

Step 3: Write the function: P(t) = 500 · 2ᵗ/³

Step 4: Verify: At t = 0: P(0) = 500 · 2⁰ = 500 ✓ At t = 3: P(3) = 500 · 2³/³ = 500 · 2 = 1000 ✓ At t = 6: P(6) = 500 · 2⁶/³ = 500 · 4 = 2000 ✓

Answer: P(t) = 500 · 2ᵗ/³

11Problem 11hard

❓ Question:

A car purchased for $25,000 depreciates at a rate of 15% per year. Write an exponential decay function V(t) for the car's value after t years, and find its value after 5 years.

💡 Show Solution

Step 1: Identify the exponential decay formula: V(t) = V₀(1 - r)ᵗ

Where:

  • V₀ = initial value
  • r = decay rate (as decimal)
  • t = time in years

Step 2: Identify the values: V₀ = 25,000 r = 0.15 (15% as decimal) 1 - r = 0.85

Step 3: Write the function: V(t) = 25,000(0.85)ᵗ

Step 4: Find value after 5 years: V(5) = 25,000(0.85)⁵

Step 5: Calculate (0.85)⁵: 0.85⁵ ≈ 0.4437

Step 6: Find the value: V(5) = 25,000 × 0.4437 V(5) ≈ $11,092.50

Step 7: Interpret: After 5 years, the car has lost about 56% of its value Original: 25,000After5years:25,000 After 5 years: 11,092.50 Lost: $13,907.50

Answer: V(t) = 25,000(0.85)ᵗ; After 5 years: approximately $11,092.50

Explain using:

⚠️ Common Mistakes: Exponential Functions

Avoid these 3 frequent errors

🌍 Real-World Applications: Exponential Functions

See how this math is used in the real world

📝 Worked Example: Solving a Quadratic by Factoring

Problem:

Solve x2−5x+6=0x^2 - 5x + 6 = 0.

2Factor the quadratic
3Set each factor equal to zero
Next Topic →
Logarithmic Functions
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📌 Related Topics in Exponential and Logarithmic Functions

Logarithmic FunctionsSolving Exponential EquationsSolving Logarithmic Equations

❓ Frequently Asked Questions

What is Exponential Functions?▾
Properties and graphs of exponential functions
How can I study Exponential Functions effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 11 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Exponential Functions study guide free?▾
Yes — all study notes, flashcards, and practice problems for Exponential Functions on Study Mondo are free to access. No account is needed.
What course covers Exponential Functions?▾
Exponential Functions is part of the Algebra 2 course on Study Mondo, specifically in the Exponential and Logarithmic Functions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Exponential Functions?▾
Yes, this page includes 11 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.