Exponential Functions and Growth/Decay

Understanding exponential functions, exponential growth, and exponential decay models

Exponential Functions and Growth/Decay

What is an Exponential Function?

An exponential function is a function of the form:

$f(x) = ab^x$

where:

$a$ is the initial value (when $x = 0$ )
$b$ is the base (must be positive and $b \neq 1$ )
$x$ is the exponent (the variable)

Key Property: The variable is in the exponent, not the base!

Basic Exponential Functions

The parent function is $f(x) = b^x$ where $b > 0$ and $b \neq 1$ .

Two Cases:

Exponential Growth when $b > 1$
- Example: $f(x) = 2^x$
- The function increases as $x$ increases
- Rises steeply to the right
Exponential Decay when $0 < b < 1$
- Example: $f(x) = \left(\frac{1}{2}\right)^x$
- The function decreases as $x$ increases
- Approaches zero but never reaches it

Properties of Exponential Functions

For $f(x) = ab^x$ where $a > 0$ and $b > 0, b \neq 1$ :

Domain: All real numbers $(-\infty, \infty)$
Range: $(0, \infty)$ if $a > 0$ , or $(-\infty, 0)$ if $a < 0$
y-intercept: $(0, a)$ since $f(0) = ab^0 = a$
Horizontal Asymptote: $y = 0$ (the x-axis)
Always positive (if $a > 0$ ) - never crosses the x-axis

Exponential Growth and Decay Models

Growth Model

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

where:

$A(t)$ = amount after time $t$
$A_0$ = initial amount
$r$ = growth rate (as a decimal)
$t$ = time

Decay Model

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

where:

$r$ = decay rate (as a decimal)

Continuous Compounding

$A(t) = A_0 e^{rt}$

where $e \approx 2.71828$ (Euler's number)

Key Patterns

Doubling: If something doubles, multiply by 2
Tripling: If something triples, multiply by 3
Half-life: If something halves, multiply by $\frac{1}{2}$
Growth by 5%: Multiply by $(1 + 0.05) = 1.05$
Decay by 5%: Multiply by $(1 - 0.05) = 0.95$

Transformations

Just like other functions, exponential functions can be transformed:

$f(x) = ab^x + k$ : vertical shift by $k$
$f(x) = ab^{x-h}$ : horizontal shift by $h$
$f(x) = -ab^x$ : reflection over x-axis

📚 Practice Problems

1Problem 1medium

❓ Question:

A population of bacteria starts with 200 bacteria and doubles every 3 hours. Write an exponential function to model the population after $t$ hours.

💡 Show Solution

Solution:

Step 1: Identify the initial value and growth pattern.

Initial population: $A_0 = 200$
The population doubles every 3 hours

Step 2: Determine the base for doubling every 3 hours.

Since it doubles every 3 hours, after 3 hours we have $2 \times 200 = 400$ .

We need to find $b$ such that $200b^3 = 400$ .

$b^3 = 2$ $b = 2^{1/3} = \sqrt[3]{2}$

Step 3: Write the function.

$A(t) = 200 \cdot 2^{t/3}$

OR equivalently: $A(t) = 200 \cdot (\sqrt[3]{2})^t$

Step 4: Verify.

At $t = 0$ : $A(0) = 200 \cdot 2^{0/3} = 200$ ✓
At $t = 3$ : $A(3) = 200 \cdot 2^{3/3} = 200 \cdot 2 = 400$ ✓
At $t = 6$ : $A(6) = 200 \cdot 2^{6/3} = 200 \cdot 4 = 800$ ✓

Answer: $A(t) = 200 \cdot 2^{t/3}$ bacteria

2Problem 2easy

❓ Question:

A population of bacteria doubles every 3 hours. Initially, there are 500 bacteria.

a) Write an exponential function $P(t)$ that models the population after $t$ hours. b) How many bacteria will there be after 12 hours? c) How long will it take for the population to reach 16,000?

💡 Show Solution

Solution:

Part (a): For exponential growth with doubling, we use: $P(t) = P_0 \cdot 2^{t/T}$

where:

$P_0 = 500$ (initial population)
$T = 3$ (doubling time in hours)

$P(t) = 500 \cdot 2^{t/3}$

Part (b): After 12 hours:

$P(12) = 500 \cdot 2^{12/3} = 500 \cdot 2^4 = 500 \cdot 16 = 8000$ bacteria

Part (c): Set $P(t) = 16000$ and solve:

$16000 = 500 \cdot 2^{t/3}$

$32 = 2^{t/3}$

$2^5 = 2^{t/3}$

$5 = \frac{t}{3}$

$t = 15$ hours

3Problem 3easy

❓ Question:

A car purchased for $25,000 depreciates by 15% each year. What is the car worth after 5 years?

💡 Show Solution

Solution:

Step 1: Identify the initial value and decay rate.

Initial value: $A_0 = 25000$
Decay rate: $r = 0.15$ (15%)

Step 2: Write the exponential decay model.

Since the car loses 15% each year, it retains 85% of its value: $A(t) = 25000(1 - 0.15)^t = 25000(0.85)^t$

Step 3: Calculate the value after 5 years.

$A(5) = 25000(0.85)^5$ $A(5) = 25000(0.4437...)$ $A(5) \approx 11,092.63$

Answer: The car is worth approximately $11,093 after 5 years.

4Problem 4medium

❓ Question:

A radioactive substance decays according to the formula $A(t) = A_0 e^{-0.0315t}$ , where $t$ is in years.

a) What is the decay rate (as a percentage)? b) What is the half-life of the substance? c) If you start with 200 grams, how much will remain after 50 years?

💡 Show Solution

Solution:

Part (a): The formula $A(t) = A_0 e^{kt}$ has decay rate $|k|$ .

Here $k = -0.0315$ , so the decay rate is $0.0315 = 3.15\%$ per year.

Part (b): Half-life is when $A(t) = \frac{A_0}{2}$ :

$\frac{A_0}{2} = A_0 e^{-0.0315t}$

$\frac{1}{2} = e^{-0.0315t}$

$\ln(0.5) = -0.0315t$

$t = \frac{\ln(0.5)}{-0.0315} = \frac{-0.693}{-0.0315} \approx 22.0$ years

Part (c): With $A_0 = 200$ grams and $t = 50$ years:

$A(50) = 200 \cdot e^{-0.0315(50)}$

$A(50) = 200 \cdot e^{-1.575}$

$A(50) = 200 \cdot 0.2067$

$A(50) \approx 41.3$ grams

5Problem 5easy

❓ Question:

Determine whether each function represents exponential growth or decay: (a) $f(x) = 5(1.2)^x$ , (b) $g(x) = 3(0.8)^x$ , (c) $h(x) = 2e^{-0.5x}$

💡 Show Solution

Solution:

Part a) $f(x) = 5(1.2)^x$

Base: $b = 1.2 > 1$

Since the base is greater than 1, this represents exponential growth.

Part b) $g(x) = 3(0.8)^x$

Base: $b = 0.8 < 1$ (but $0.8 > 0$ )

Since the base is between 0 and 1, this represents exponential decay.

Part c) $h(x) = 2e^{-0.5x}$

Rewrite: $h(x) = 2(e^{-0.5})^x$

The base is $e^{-0.5} \approx 0.606 < 1$

Since the base is between 0 and 1, this represents exponential decay.

Alternative approach for part c): The negative exponent indicates decay.

Answers:

a) Growth
b) Decay
c) Decay

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