🎯⭐ INTERACTIVE LESSON

Exponential Functions

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Exponential Functions - Complete Interactive Lesson

Part 1: Exponential Growth & Decay

📈 Exponential Functions — Core Form & Growth vs Decay

Part 1 of 7

An exponential function has the form:

$\boxed{f(x) = a \cdot b^x, \quad a \neq 0,\; b > 0,\; b \neq 1}$

Parameter	Name	What It Controls
$a$	Initial value	$y$ -intercept: $f(0) = a$
$b$

Key insight: Every time $x$ increases by $1$ , the output is multiplied by $b$ — not added to. This is what separates exponential from linear.

🔍 Growth vs Decay — Side by Side

Growth: $f(x) = 3 \cdot 2^x$

$x$	$f(x)$

💹 Percent Growth & Decay Rate

In applications, the base $b$ is often written as:

$\boxed{b = 1 + r \quad \text{(growth)} \qquad b = 1 - r \quad \text{(decay)}}$

Concept Check 🎯

Calculation Drill 🧮

1) A bacteria culture starts with $200$ cells and triples every hour. How many cells after $4$ hours? (e.g., doubling: $50 \cdot 2^3 = 400$ )

2) An investment of $\$ 1{,}0005%1$80010%800(1.10) = 880$)

Classify Each Scenario 🔽

Exit Quiz ✅

Part 2: Properties of Exponential Functions

🔄 Transformations of Exponential Graphs

Part 2 of 7

The parent exponential function is $f(x) = b^x$ . Every transformation follows the general form:

$\boxed{g(x) = a \cdot b^{x - h} + k}$

Part 3: Transformations

💰 Compound Interest & Continuous Growth

Part 3 of 7

When interest is compounded periodically, we use:

$\boxed{A = P\left(1 + \frac{r}{n}\right)^{nt}}$

Part 4: Real-World Models

☢️ Half-Life & Exponential Decay

Part 4 of 7

Half-life is the time it takes for a quantity to reduce to half its current value.

$\boxed{A(t) = A_0 \cdot \left(\frac{1}{2}\right)^{t/T_{1/2}}}$

Part 5: Compound Interest & e

🔑 Solving Exponential Equations with Logarithms

Part 5 of 7

When the variable is in the exponent, logarithms are the key tool.

The Core Technique

$b^x = c \quad \Longrightarrow \quad \boxed{x = \frac{\ln c}{\ln b} = \log_b c}$

Part 6: Problem-Solving Workshop

📊 Data Fitting & Exponential Regression

Part 6 of 7

How do you determine whether data is exponential — and if so, find the model $f(x) = ab^x$ ?

The Ratio Test

If data is exponential, consecutive $y$ -values have a constant ratio.

$x$

Part 7: Review & Applications

🏆 Exponential Functions — Full Synthesis

Part 7 of 7 — Putting It All Together

This final part combines every exponential skill into multi-step problems — just like the AP exam.

Your Exponential Toolkit

Concept (Part)	Key Formula / Idea
Core Form (1)	$f(x) = ab^x$ , growth if $b > 1$ , decay if

Consecutive outputs always have the same ratio ( $b = 2$ ). This constant ratio is the hallmark of exponential behavior.

Decay: $g(x) = 80 \cdot (0.5)^x$

$x$	$g(x)$	Ratio
$0$	$80$	—
$1$	$40$	$0.5$
$2$	$20$	$0.5$
$3$	$10$	$0.5$

Each step halves the output. The base $b = 0.5$ is between $0$ and $1$ , so the function decays.

📐 Graph Features (Both Cases)

Feature	Value
Domain	All real numbers $(-\infty, \infty)$
Range	$(0, \infty)$ when $a > 0$
$y$ -intercept	$(0, a)$
Horizontal asymptote	$y = 0$ (the $x$ -axis)
Passes through	$(1, ab)$ always

where $r$ is the percent rate (as a decimal).

A town of $5{,}000$ people grows $6\%$ per year. Write the model and find the population after $10$ years.

$P(t) = 5000(1 + 0.06)^t = 5000(1.06)^t$

$P(10) = 5000(1.06)^{10} = 5000(1.7908...) \approx 8{,}954$

Scenario	$r$	$b = 1 \pm r$	Growth or Decay?
$15\%$ annual appreciation	$0.15$	$1.15$	Growth
$8\%$ annual depreciation	$0.08$	$0.92$	Decay
$3.5\%$ monthly increase	$0.035$	$1.035$	Growth
$20\%$ hourly radioactive loss	$0.20$	$0.80$	Decay

annually. What is it worth after

3) Evaluate $6 \cdot \left(\frac{1}{3}\right)^2$ . (e.g., $4 \cdot \left(\frac{1}{2}\right)^3 = 4 \cdot \frac{1}{8} = 0.5$ )

Parameter	Effect	Example on $2^x$
$a$	Vertical stretch/compress; if $a < 0$ , reflect over $x$ -axis	$-2^x$ flips graph upside down
$h$	Horizontal shift (right if $h > 0$ )	$2^{x-3}$ shifts right $3$
$k$	Vertical shift (up if $k > 0$ )	$2^x + 5$ shifts up $5$
Replace $x$ with $-x$	Reflect over $y$ -axis	$2^{-x} = \left(\frac{1}{2}\right)^x$

Critical change: The horizontal asymptote moves from $y = 0$ to $y = k$ whenever a vertical shift is applied.

🧩 Transformation Breakdown

Vertical Stretch & Reflection ( $a$ )

Function	$a$ value	Effect
$3 \cdot 2^x$	$a = 3$	Stretched vertically by factor $3$ ; $y$ -int at $(0, 3)$
$0.5 \cdot 2^x$	$a = 0.5$	Compressed vertically; $y$ -int at $(0, 0.5)$
$-2^x$	$a = -1$	Reflected over $x$ -axis; range becomes $(-\infty, 0)$

Horizontal & Vertical Shifts ( $h$ and $k$ )

Function	Shift	New HA	New $y$ -intercept
$2^{x-2}$	Right $2$	$y = 0$

Worked Example

Graph $g(x) = -3 \cdot 2^{x+1} + 4$ starting from the parent $f(x) = 2^x$ .

Step	Transformation	Key Point $(0,1) \to$	HA
Start	$2^x$	$(0, 1)$

Final: HA at $y = 4$ , $y$ -intercept at $g(0) = -3(2) + 4 = -2$ , range .

📐 Domain & Range After Transformations

The domain of exponential functions is always $(-\infty, \infty)$ — no transformation changes this.

The range depends on $a$ and $k$ :

Condition	Range
$a > 0$	$(k, \infty)$
$a < 0$	$(-\infty, k)$

Quick Check Method

To find the $y$ -intercept of $g(x) = a \cdot b^{x-h} + k$ :

$\boxed{g(0) = a \cdot b^{-h} + k}$

To find where $g(x) = 0$ (if it crosses the $x$ -axis):

Set $a \cdot b^{x-h} + k = 0 \implies b^{x-h} = -\frac{k}{a}$

This has a solution only when $-\frac{k}{a} > 0$ (i.e., $k$ and $a$ have opposite signs).

Transformation Check 🎯

Transformation Drill 🧮

1) Find the $y$ -intercept of $g(x) = 6 \cdot 3^{x-1} - 2$ . (e.g., for $h(x) = 4 \cdot 2^{x-2} + 1$ : $h(0) = 4 \cdot 2^{-2} + 1 = 4 \cdot 0.25 + 1 = 2$ )

2) What is the horizontal asymptote ( $y$ -value) of $f(x) = -10 \cdot 5^x + 12$ ? (e.g., $3 \cdot 2^{x} -$ has HA )

3) The function $f(x) = 2^{x+3}$ is equivalent to $c \cdot 2^x$ . What is ? (e.g., so )

Identify the Transformation 🔽

Variable	Meaning
$A$	Final amount
$P$	Principal (initial investment)
$r$	Annual interest rate (decimal)
$n$	Number of compounding periods per year
$t$	Time in years

Common Compounding Frequencies

Frequency	$n$
Annually	$1$
Semi-annually	$2$
Quarterly	$4$
Monthly	$12$
Daily	$365$
Continuously	Use $A = Pe^{rt}$ instead

📊 Worked Examples

Example 1: Quarterly Compounding

$\$ 5{,}000 $is invested at$ 6% $annual interest compounded quarterly. Find the balance after$ 3$ years.

$A = 5000\left(1 + \frac{0.06}{4}\right)^{4 \cdot 3} = 5000(1.015)^{12}$

$(1.015)^{12} \approx 1.19562$ → $A \approx \$ 5{,}978.09$

Example 2: Comparing Frequencies

$\$ 10{,}000 $at$ 8% $for$ 5$ years. Compare annual vs monthly compounding.

Frequency	$n$	Calculation	Final Amount
Annual	$1$	$10000(1.08)^5$	$\$ 14{,}693.28$
Monthly

More frequent compounding always gives a higher return, but with diminishing marginal benefit.

♾️ Continuous Compounding & the Number $e$

As $n \to \infty$ , the compound interest formula approaches:

$\boxed{A = Pe^{rt}}$

where $e \approx 2.71828...$

Why $e$ ?

$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$

This limit is the foundation of continuous growth.

Worked Example

$\$ 2{,}000 $invested at$ 5% $compounded continuously for$ 10$ years.

$A = 2000e^{0.05 \cdot 10} = 2000e^{0.5} = 2000(1.6487...) \approx \$ 3{,}297.44$

Converting Between Forms

To convert $A = Pe^{rt}$ to $A = P \cdot b^t$ :

$b = e^r$

To convert $A = P \cdot b^t$ to $A = Pe^{rt}$ :

$r = \ln(b)$

Periodic Form	Continuous Equivalent
$1000(1.06)^t$	$1000e^{0.0583t}$ because $\ln(1.06) \approx 0.0583$

Compound Interest Check 🎯

Compound Interest Drill 🧮

1) $\$ 1{,}000 $at$ 10% $compounded annually for$ 2 $years. What is$ A $? (e.g.,$ $500 $at$ 8% $annually for$ 1 $year:$ 500(1.08) = 540$)

2) How many compounding periods in $4$ years of monthly compounding? (e.g., quarterly for $3$ years: $4 \times 3 = 12$ periods)

3) Evaluate $e^1$ rounded to two decimal places. (e.g., $e^0 = 1.00$ )

Classify Each Scenario 🔽

Variable	Meaning
$A(t)$	Amount remaining at time $t$
$A_0$	Initial amount
$T_{1/2}$	Half-life (time to halve)

This formula works because after each half-life period:

After $1$ half-life: $\frac{1}{2}$ remains
After $2$ half-lives: $\frac{1}{4}$ remains
After $3$ half-lives: $\frac{1}{8}$ remains
After $n$ half-lives: $\left(\frac{1}{2}\right)^n$ remains

🧪 Worked Examples

Example 1: Carbon-14 Dating

Carbon-14 has a half-life of $5{,}730$ years. A fossil has $25\%$ of its original C-14. How old is it?

$25\% = \frac{1}{4} = \left(\frac{1}{2}\right)^2$ → exactly $2$ half-lives have passed.

Age $= 2 \times 5730 = 11{,}460$ years.

Example 2: Medicine Clearance

A drug has a half-life of $4$ hours. A patient takes $200\text{ mg}$ . How much remains after $10$ hours?

$A(10) = 200 \cdot \left(\frac{1}{2}\right)^{10/4} = 200 \cdot \left(\frac{1}{2}\right)^{2.5}$

$= 200 \cdot \frac{1}{2^{2.5}} = 200 \cdot \frac{1}{5.657} \approx 35.4\text{ mg}$

Reference Table: Common Real-World Half-Lives

Substance	Half-Life	Context
Carbon-14	$5{,}730$ years	Archaeological dating
Iodine-131	$8$ days	Thyroid treatment
Caffeine	$\sim 5$ hours	Metabolism
Uranium-238	$4.5$ billion years

🔗 Connecting Half-Life to the Decay Constant

The continuous decay model $A(t) = A_0 e^{kt}$ (with $k < 0$ ) is related to half-life by:

$\boxed{T_{1/2} = \frac{\ln 2}{|k|} \quad \Longleftrightarrow \quad k = -\frac{\ln 2}{T_{1/2}}}$

Worked Example

A radioactive sample decays according to $A(t) = 500e^{-0.1t}$ (grams, hours). Find the half-life.

$T_{1/2} = \frac{\ln 2}{|-0.1|} = \frac{0.6931}{0.1} \approx 6.93 \text{ hours}$

Converting Between Forms

Given	Find	Method
Half-life $= 8$ days	Decay constant $k$	$k = -\frac{\ln 2}{8} \approx -0.0866$

Half-Life Check 🎯

Decay Calculations 🧮

1) A sample starts at $400\text{ g}$ with half-life $5$ years. How many grams remain after $15$ years? (e.g., $600\text{ g}$ with half-life $3$ years after $6$ years: $600 \cdot (\frac{1}{2})^2 = 150\text{ g}$ )

2) If $k = -0.04$ , what is the half-life? Round to one decimal. (e.g., $k = -0.1$ : $T_{1/2} = \frac{0.693}{0.1} = 6.9$ )

3) How many half-lives occur in $24$ hours if each half-life is $8$ hours? (e.g., half-life of $6$ hours in $18$ hours: $18/6 = 3$ )

Decay Concepts 🔽

Strategy	When to Use	Example
Same-base matching	Both sides are powers of the same base	$4^x = 8$ → $2^{2x} = 2^3$
Take $\ln$ of both sides	Bases can't be matched easily	$5^x = 20$ → $x = \frac{\ln 20}{\ln 5}$
Change of base formula	Need a decimal approximation	$\log_3 7 = \frac{\ln 7}{\ln 3} \approx 1.771$

🔗 Same-Base Method

Worked Example 1

Solve $8^{2x+1} = 32$ .

Rewrite as powers of $2$ : $(2^3)^{2x+1} = 2^5$

$2^{3(2x+1)} = 2^5$

$3(2x + 1) = 5$

$6x + 3 = 5 \implies 6x = 2 \implies x = \frac{1}{3}$

Common Base Conversions

Number	As power of $2$	As power of $3$
$4$	$2^2$	—
$8$

📐 Logarithm Method (General Case)

Worked Example 2

Solve $3 \cdot 5^{x-1} = 90$ .

Step 1 — Isolate the exponential: $5^{x-1} = 30$

Step 2 — Take $\ln$ of both sides: $(x - 1)\ln 5 = \ln 30$

Step 3 — Solve for $x$ : $x - 1 = \frac{\ln 30}{\ln 5} = \frac{3.401}{1.609} \approx 2.113$

$x \approx 3.113$

Worked Example 3: Application

A population of $2{,}000$ grows at $4\%$ per year. When will it reach $5{,}000$ ?

$2000(1.04)^t = 5000$

$(1.04)^t = 2.5$

$t = \frac{\ln 2.5}{\ln 1.04} = \frac{0.9163}{0.03922} \approx 23.4$ years

⚠️ Common Error

Never distribute $\ln$ across addition: $\ln(a + b) \neq \ln a + \ln b$ . Logarithms only split across products and quotients: .

Equation-Solving Check 🎯

Solve for $x$ 🧮

1) $4^x = 64$ . (e.g., $3^x = 81$ : since $81 = 3^4$ , $x = 4$ )

2) $e^{2x} = 7.389$ . Round to one decimal. (e.g., $e^x = 2.718$ : $x = \ln(2.718) \approx 1.0$ )

3) $\log_2 32 = ?$ (e.g., $\log_3 27 = 3$ because )

Strategy Selection 🔽

Constant ratio $= 3$ → exponential with $b = 3$ .

Linear data has constant differences. Exponential data has constant ratios.

🔍 Finding $a$ and $b$ From Data

Method 1: Two Points

Given two points $(x_1, y_1)$ and $(x_2, y_2)$ on $f(x) = ab^x$ :

$\boxed{b = \left(\frac{y_2}{y_1}\right)^{\frac{1}{x_2 - x_1}}} \qquad \boxed{a = \frac{y_1}{b^{x_1}}}$

Worked Example

Find the exponential function through $(2, 12)$ and $(5, 324)$ .

Step 1 — Find $b$ : $b = \left(\frac{324}{12}\right)^{1/(5-2)} = 27^{1/3} = 3$

Step 2 — Find $a$ : $a = \frac{12}{3^2} = \frac{12}{9} = \frac{4}{3}$

Result: $f(x) = \frac{4}{3} \cdot 3^x$

Verify: $f(5) = \frac{4}{3} \cdot 243 = 324$ ✔

📈 Log Linearization

Taking logarithms converts exponential data into linear data:

$y = ab^x \quad \Longrightarrow \quad \ln y = \ln a + x \ln b$

This is the form $Y = mx + c$ where:

$Y = \ln y$
$m = \ln b$ (slope)
$c = \ln a$ ( $y$ -intercept)

Why This Matters

Raw Data Plot	Log-Transformed Plot
Curved (exponential shape)	Straight line
Hard to determine $a$ and $b$ visually	Slope gives $\ln b$ , intercept gives $\ln a$

Worked Example

Data: $(0, 2)$ , $(1, 6)$ , $(2, 18)$ , $(3, 54)$ . Verify exponential and find the model using logs.

$x$	$y$	$\ln y$
$0$	$2$

$\ln y$ values have constant spacing ( $\approx 1.099$ ) → confirms exponential.

Slope $= 1.099 = \ln b \implies b = e^{1.099} \approx 3$

Intercept $= 0.693 = \ln a \implies a = e^{0.693} = 2$

Model: $f(x) = 2 \cdot 3^x$

Data Analysis Check 🎯

Data Fitting Drill 🧮

1) Data: $(0, 10)$ , $(1, 20)$ , $(2, 40)$ . What is $b$ ? (e.g., for data $5, 15, 45$ : ratio $= 15/5 = 3$ , so $b = 3$ )

2) For the model $y = 5 \cdot 2^x$ , what is $\ln(a)$ ? Round to two decimal places. (e.g., if $a = 3$ : )

3) An exponential function passes through $(0, 8)$ and $(3, 64)$ . What is $b$ ? (e.g., through $(0, 5)$ and : )

Data Interpretation 🔽

📋 Multi-Step Problem Walkthrough

A pharmaceutical company tests a new drug. At $t = 0$ , the bloodstream concentration is $200\text{ mg/L}$ . After $3$ hours, it's $50\text{ mg/L}$ .

(a) Find the exponential model. (b) Find the half-life. (c) When does concentration drop below $5\text{ mg/L}$ ?

Part (a): Find the model

$C(t) = 200 \cdot b^t$

Using $C(3) = 50$ : $200b^3 = 50 \implies b^3 = 0.25 \implies b = 0.25^{1/3} \approx 0.630$

$\boxed{C(t) = 200(0.630)^t}$

Part (b): Find the half-life

$T_{1/2} = \frac{\ln 2}{|\ln(0.630)|} = \frac{0.693}{0.462} \approx 1.5\text{ hours}$

Check: $C(1.5) = 200(0.630)^{1.5} = 200(0.5) = 100$ ✔ (half of $200$ )

Part (c): When $C(t) < 5$ ?

$200(0.630)^t = 5$

$(0.630)^t = 0.025$

$t = \frac{\ln(0.025)}{\ln(0.630)} = \frac{-3.689}{-0.462} \approx 7.98\text{ hours}$

Concentration drops below $5\text{ mg/L}$ after about $8$ hours.

⚖️ Comparing Exponential Models

When Problems Give Different Formats

Given Format	Convert To	Method
"Doubles every $T$ years"	$f(t) = a \cdot 2^{t/T}$	Base is $2$ , exponent is $t/T$
"Grows $r\%$ per year"	$f(t) = a(1 + r)^t$	Standard percent form
"Continuous rate $k$ "	$f(t) = ae^{kt}$	Natural exponential
"Half-life of $T$ hours"	$f(t) = a(\frac{1}{2})^{t/T}$

Comparison Example

Investment A: $\$ 10{,}000 $at$ 5% $compounded annually. Investment B:$ $8{,}0006%$ compounded continuously. When does B overtake A?

$A(t) = 10000(1.05)^t$

$B(t) = 8000e^{0.06t}$

Set equal: $10000(1.05)^t = 8000e^{0.06t}$

$\frac{10000}{8000} = \frac{e^{0.06t}}{(1.05)^t} = \left(\frac{e^{0.06}}{1.05}\right)^t$

$1.25 = (1.01184)^t \implies t = \frac{\ln 1.25}{\ln 1.01184} \approx \frac{0.2231}{0.01177} \approx 18.9$ years

Synthesis Quiz 🎯

Multi-Step Drill 🧮

1) A substance has half-life $6$ hours. Starting from $480\text{ g}$ , how many grams remain after $18$ hours? (e.g., half-life $4$ hours, start $320\text{ g}$ , after $12$ hours: $320 \cdot (\frac{1}{2})^3 = 40\text{ g}$ )

2) Solve $3 \cdot 2^x = 96$ . (e.g., $5 \cdot 3^x = 405$ : divide by to get , so )

3) An account at $8\%$ compounded annually doubles when $t = \frac{\ln 2}{\ln 1.08}$ . Round to one decimal. (e.g., at : years)

Formula Selection 🔽

Final Exit Quiz — Exponential Functions ✅

3^{x+2} = 9 \cdot 3^x

ln (1.06) \approx 0.0583

ln (ab) = ln a + ln b

x = ln (2.718) \approx 1.0

\ln 3 \approx 1.10

b = \sqrt{45/5} = 3

\frac{0.693}{0.0953} \approx 7.3

Exponential Functions

Exponential Functions - Complete Interactive Lesson

Part 1: Exponential Growth & Decay

📈 Exponential Functions — Core Form & Growth vs Decay

🔍 Growth vs Decay — Side by Side

Growth: f(x)=3⋅2xf(x) = 3 \cdot 2^xf(x)=3⋅2x

💹 Percent Growth & Decay Rate

Part 2: Properties of Exponential Functions

🔄 Transformations of Exponential Graphs

Part 3: Transformations

💰 Compound Interest & Continuous Growth

Part 4: Real-World Models

☢️ Half-Life & Exponential Decay

Part 5: Compound Interest & e

🔑 Solving Exponential Equations with Logarithms

The Core Technique

Part 6: Problem-Solving Workshop

📊 Data Fitting & Exponential Regression

The Ratio Test

Part 7: Review & Applications

🏆 Exponential Functions — Full Synthesis

Your Exponential Toolkit

Decay: g(x)=80⋅(0.5)xg(x) = 80 \cdot (0.5)^xg(x)=80⋅(0.5)x

📐 Graph Features (Both Cases)

Worked Example

Quick Reference

🧩 Transformation Breakdown

Vertical Stretch & Reflection (aaa)

Horizontal & Vertical Shifts (hhh and kkk)

Worked Example

📐 Domain & Range After Transformations

Quick Check Method

Common Compounding Frequencies

📊 Worked Examples

Example 1: Quarterly Compounding

Example 2: Comparing Frequencies

♾️ Continuous Compounding & the Number eee

Why eee?

Worked Example

Converting Between Forms

🧪 Worked Examples

Example 1: Carbon-14 Dating

Example 2: Medicine Clearance

Reference Table: Common Real-World Half-Lives

🔗 Connecting Half-Life to the Decay Constant

Worked Example

Converting Between Forms

🔗 Same-Base Method

Worked Example 1

Common Base Conversions

📐 Logarithm Method (General Case)

Worked Example 2

Worked Example 3: Application

⚠️ Common Error

🔍 Finding aaa and bbb From Data

Method 1: Two Points

Worked Example

📈 Log Linearization

Why This Matters

Worked Example

📋 Multi-Step Problem Walkthrough

Part (a): Find the model

Part (b): Find the half-life

Part (c): When C(t)<5C(t) < 5C(t)<5?

⚖️ Comparing Exponential Models

When Problems Give Different Formats

Comparison Example

Growth: $f(x) = 3 \cdot 2^x$

Decay: $g(x) = 80 \cdot (0.5)^x$

Vertical Stretch & Reflection ( $a$ )

Horizontal & Vertical Shifts ( $h$ and $k$ )

♾️ Continuous Compounding & the Number $e$

Why $e$ ?

🔍 Finding $a$ and $b$ From Data

Part (c): When $C(t) < 5$ ?