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🎯⭐ INTERACTIVE LESSON

Exponential Functions

Learn step-by-step with interactive practice!

Exponential Functions - Complete Interactive Lesson

Part 1: Exponential Growth

📐 Exponential Growth

Part 1 of 7 — Exponential Growth

y=a⋅bxquad(b>1)y = a \cdot b^x quad (b > 1)

  • aa = initial value
  • bb = growth factor
  • Growth rate rr: b=1+rb = 1 + r

Exponential growth: each step multiplies by bb.

Worked Example

Population: 500, grows 10% per year. After 3 years?

y=500(1.10)3=500(1.331)=665.5y = 500(1.10)^3 = 500(1.331) = 665.5 ✅

Concept Check 🎯

Exponential Growth 🧮

  1. 100(1.2)2=?100(1.2)^2 = ?

  2. 1000(1.5)1=?1000(1.5)^1 = ?

  3. y=2xy = 2^x. When x=3x = 3, y=?y = ?

Concept Check 🔍

Practice

#InitialRateYearsFormula
110020%2100(1.2)2100(1.2)^{2}
22005%3200(1.05)3200(1.05)^{3}
3100050%11000(1.5)11000(1.5)^{1}

Challenge Question 📋

Part 2: Exponential Decay

📊 Exponential Decay

Part 2 of 7 — Exponential Decay

y=a⋅bxquad(0<b<1)y = a \cdot b^x quad (0 < b < 1)

Decay factor: b=1−rb = 1 - r (where rr is the decay rate)

Half-life: time for the quantity to halve.

Worked Example

Car: $20,000, depreciates 15%/year. Value after 2 years?

y=20000(0.85)2=20000(0.7225)=14450y = 20000(0.85)^2 = 20000(0.7225) = 14450 → $14,450 ✅

Concept Check 🎯

Exponential Decay 🧮

  1. 1000(0.9)2=?1000(0.9)^2 = ?

  2. 500(0.5)1=?500(0.5)^1 = ?

  3. 800(0.75)2=?800(0.75)^2 = ?

Concept Check 🔍

Practice

#InitialRateYearsFormula
1100010%21000(0.9)21000(0.9)^{2}
250050%1500(0.5)1500(0.5)^{1}
380025%2800(0.75)2800(0.75)^{2}

Challenge Question 📋

Part 3: Compound Interest

🔢 Compound Interest

Part 3 of 7 — Compound Interest

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

  • PP = principal, rr = annual rate, nn = compounding frequency, tt = years

Continuous compounding: A=PertA = Pe^{rt}

Worked Example

$1000 at 6% compounded annually for 2 years.

A=1000(1.06)2=1000(1.1236)=1123.6A = 1000(1.06)^2 = 1000(1.1236) = 1123.6 → $1,123.60 ✅

Concept Check 🎯

Compound Interest 🧮

  1. $500 at 10%, 1 year, annual. A = ?

  2. $1000 at 5%, 2 years, annual. A = ?

  3. $1000 at 6%, 1 year, annual. A = ?

Concept Check 🔍

Practice

#PrntA
1$10005%12$1102.50
2$50010%11$550
3$20004%13$2249.73

Challenge Question 📋

Part 4: Logarithms Introduction

📈 Logarithms Introduction

Part 4 of 7 — Logarithms Introduction

A logarithm is the inverse of an exponential.

logb(x)=yiffby=xlog_b(x) = y iff b^y = x

log⁡2(8)=3\log_2(8) = 3 because 23=82^3 = 8

Common log: log⁡(x)=log10(x)\log(x) = log_{10}(x) Natural log: ln(x)=loge(x)ln(x) = log_e(x)

Worked Example

log3(81)=?log_3(81) = ?

3?=813^? = 81. Since 34=813^4 = 81, log3(81)=4log_3(81) = 4 ✅

Concept Check 🎯

Evaluate Logs 🧮

  1. log2(16)=?log_2(16) = ?

  2. log5(25)=?log_5(25) = ?

  3. log10(1000)=?log_{10}(1000) = ?

Concept Check 🔍

Practice

#LogValue
1log2(16)log_2(16)4
2log5(25)log_5(25)2
3log10(1000)log_{10}(1000)3

Challenge Question 📋

Part 5: Log Properties

🧮 Log Properties

Part 5 of 7 — Log Properties

Key Properties

  1. Product: logb(MN)=logb(M)+logb(N)log_b(MN) = log_b(M) + log_b(N)
  2. Quotient: logb(MN)=logb(M)−logb(N)log_b\left(\frac{M}{N}\right) = log_b(M) - log_b(N)
  3. Power: logb(Mp)=p⋅logb(M)log_b(M^p) = p \cdot log_b(M)

Worked Example

log2(4⋅8)=log2(4)+log2(8)=2+3=5log_2(4 \cdot 8) = log_2(4) + log_2(8) = 2 + 3 = 5 ✅

Check: 4×8=32=254 \times 8 = 32 = 2^5 ✅

Concept Check 🎯

Log Properties 🧮

  1. log2(4)+log2(8)=?log_2(4) + log_2(8) = ?

  2. log10(100)−log10(10)=?log_{10}(100) - log_{10}(10) = ?

  3. log10(103)=?log_{10}(10^3) = ?

Concept Check 🔍

Practice

#PropertyExample
1Productlog⁡(2)+log⁡(5)=log⁡(10)=1\log(2)+\log(5) = \log(10) = 1
2Quotientlog⁡(100)−log⁡(10)=log⁡(10)=1\log(100)-\log(10) = \log(10) = 1
3Powerlog⁡(103)=3\log(10^3) = 3

Challenge Question 📋

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Apply exponential and logarithmic skills:

  • Doubling time problems
  • Real-world decay (depreciation, radioactive)
  • Solving exponential equations with logs

Worked Example

$1000 at 8% annual. When does it double? (Rule of 72)

72÷8=972 \div 8 = 9 years (approximately) ✅

Concept Check 🎯

Applications 🧮

  1. Rule of 72: doubling at 6%. Years?

  2. 1000(1.05)2=?1000(1.05)^2 = ?

  3. log2(64)=?log_2(64) = ?

Concept Check 🔍

Practice

#ProblemAnswer
1Doubling at 6%?~12 years
2$1000 ×(1.05)²1102.5
3log⁡2(64)\log_{2}(64)6

Challenge Question 📋

Part 7: Review & Applications

🏆 Review & Applications

Part 7 of 7 — Review & Applications

Key Formulas

  • Growth: y=a(1+r)ty = a(1+r)^t, Decay: y=a(1−r)ty = a(1-r)^t
  • Compound interest: A=P(1+r/n)ntA = P(1+r/n)^{nt}
  • logb(x)=yiffby=xlog_b(x) = y iff b^y = x
  • Product/Quotient/Power rules for logs

Worked Example

log3(27)+log3(9)=3+2=5=log3(243)log_3(27) + log_3(9) = 3 + 2 = 5 = log_3(243) ✅

Concept Check 🎯

Review 🧮

  1. 500(1.1)2=?500(1.1)^2 = ?

  2. 1000(0.5)3=?1000(0.5)^3 = ?

  3. log3(27)=?log_3(27) = ?

Concept Check 🔍

Practice

#TypeProblem
1Growth500(1.1)2500(1.1)^{2}
2Decay1000(0.5)31000(0.5)^{3}
3Loglog⁡3(27)\log_{3}(27)

Challenge Question 📋