Exponential Functions

Properties and graphs of exponential functions

Exponential Functions

Definition

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

where:

$a$ = initial value (y-intercept when $x = 0$ )
$b$ = base (growth/decay factor)
$x$ = exponent (input variable)

Growth vs. Decay

Exponential Growth: $b > 1$

Function increases
Example: $f(x) = 2^x$

Exponential Decay: $0 < b < 1$

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

Properties

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

Exponential Growth/Decay Formula

$A = A_0(1 + r)^t$

where:

$A$ = final amount
$A_0$ = initial amount
$r$ = rate (as decimal)
$t$ = time

Growth: $r > 0$ (add) Decay: $r < 0$ (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: $f(x) = 3 \cdot 2^x$ when $x = 4$

💡 Show Solution

Substitute $x = 4$ into the function:

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

Answer: $f(4) = 48$

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 4medium

❓ 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 $\frac{12 \text{ hours}}{3 \text{ hours/doubling}} = 4 \text{ doublings}$

Step 2: Use the formula $A = A_0 \cdot 2^n$ $A = 500 \cdot 2^4$ $= 500 \cdot 16$ $= 8000$

Answer: 8,000 bacteria

5Problem 5medium

❓ 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

6Problem 6medium

❓ 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

7Problem 7hard

❓ 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: $A = A_0(1 - r)^t$

Given:

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

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

Answer: Approximately $11,093

8Problem 8hard

❓ 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: $25,000 After 5 years:$ 11,092.50 Lost: $13,907.50

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

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