Exponential Functions

📈 Exponential Functions — Core Form & Growth vs Decay

Part 1 of 7

An exponential function has the form:

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

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

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

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

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

📐 Graph Features (Both Cases)

💹 Percent Growth & Decay Rate

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

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

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

Worked Example

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

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

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

Quick Reference

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 $1,000 earns $5\%$ annually. What is it worth after $1$ year? (e.g., $800 at $10\%$: $800(1.10) = 880$)

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

Classify Each Scenario 🔽

Exit Quiz ✅

🔄 Transformations of Exponential Graphs

Part 2 of 7

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

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

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

🧩 Transformation Breakdown

Vertical Stretch & Reflection ($a$)

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

Worked Example

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

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

📐 Domain & Range After Transformations

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

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

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., $3 \cdot 2^x - 8$ has HA $y = -8$)

3) The function $f(x) = 2^{x+3}$ is equivalent to $c \cdot 2^x$. What is $c$? (e.g., $3^{x+2} = 9 \cdot 3^x$ so $c = 9$)

Identify the Transformation 🔽

Exit Quiz ✅

💰 Compound Interest & Continuous Growth

Part 3 of 7

When interest is compounded periodically, we use:

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

Common Compounding Frequencies

📊 Worked Examples

Example 1: Quarterly Compounding

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

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

$(1.015)^{12} \approx 1.19562$ → $A \approx 5978.09$, i.e. $5,978.09

Example 2: Comparing Frequencies

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

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.

$A = 2000e^{0.05 \cdot 10} = 2000e^{0.5} = 2000(1.6487...) \approx 3297.44$, i.e. $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 $A = Pe^{rt}$:

$r = \ln(b)$

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 🔽

Exit Quiz ✅

☢️ 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.

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

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

🔗 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

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 🔽

Exit Quiz ✅

🔑 Solving Exponential Equations with Logarithms

Part 5 of 7

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

The Core Technique

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

🔗 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

📐 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: $\ln(ab) = \ln a + \ln b$.

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 $3^3 = 27$)

Strategy Selection 🔽

Exit Quiz ✅

📊 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.

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

Worked Example

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

$\ln y$ 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 $a = 3$: $\ln 3 \approx 1.10$)

3) An exponential function passes through $(0, 8)$ and $(3, 64)$. What is $b$? (e.g., through $(0, 5)$ and $(2, 45)$: $b = \sqrt{45/5} = 3$)

Data Interpretation 🔽

Exit Quiz ✅

🏆 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

📋 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

Comparison Example

Investment A: $10,000 at $5\%$ compounded annually. Investment B: $8,000 at $6\%$ 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 $5$ to get $3^x = 81$, so $x = 4$)

3) An account at $8\%$ compounded annually doubles when $t = \frac{\ln 2}{\ln 1.08}$. Round to one decimal. (e.g., at $10\%$: $\frac{0.693}{0.0953} \approx 7.3$ years)

Formula Selection 🔽

Final Exit Quiz — Exponential Functions ✅