Week 15 — Lecture Outline · Exponential & Logarithmic Equations & Applications

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objectives covered: Objective 8 — Analyze and apply exponential and logarithmic functions, including solving equations and modeling real-world growth, decay, and financial scenarios.
SLOs touched: A (apply procedures accurately) · B (connect symbolic/numerical representations; interpret in context)
Meeting pattern: 2 sessions × 75 min = 150 min. Segment minutes below total ~150; scale to your own pattern.

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Week at a Glance

The week's big question	"If you know the rule for how something grows or shrinks, how do you find out when it hits a target — and how do you make sure your answer is actually valid?"
By the end of the week, students can…	(1) solve exponential equations by the same-base method (match bases, equate exponents); (2) solve exponential equations by taking a logarithm of both sides (apply log/ln, pull down the exponent, isolate the variable); (3) solve logarithmic equations by condensing to a single log and converting to exponential form, then checking for extraneous solutions (negative or zero log arguments); (4) set up, solve, and interpret real-world applications — compound interest for time, doubling time, half-life.
Key vocabulary	exponential equation, same-base method, logarithm of both sides, change-of-base, logarithmic equation, condensing logarithms, extraneous solution, compound interest, continuous compounding, doubling time, half-life, growth/decay model
Materials	slides (Deck 15), readings + video links, a calculator, Desmos (or GeoGebra) for graphical confirmation, one approved chatbot (Gemini / Claude / ChatGPT) for the AI-critique moment and the tutorial
Timing note	8 segments, ~150 min total. Session 1 = Segments 1–4 (~73 min). Session 2 = Segments 5–8 (~77 min).

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Segment 1 — Hook & the Promise (8 min) · Session 1 opens

Hook. "Raise your hand if you've ever wondered how long it takes money to double in a savings account, or how long a pain reliever stays in your system, or when a population doubles." Wait — hands go up. "Every one of those is the same algebra question: you know the rule for how something changes, and you need to find the time. That is exactly what we solve this week."

The promise (write it on the board): "By Thursday, you'll be able to take any 'how long until?' question — money, drugs, populations, radioactivity — write the equation, solve it, and give a real answer with real units."

Callback to Weeks 13–14: "We spent two weeks learning what exponential and logarithmic functions look like and how they're related. This week is the payoff: we flip those relationships into equations and solve for the unknown in the exponent — the one place ordinary algebra can't reach without logarithms."

Why it matters line: "The 'solve for t' skill is the bridge between the math and the answer the world actually cares about."

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Segment 2 — Method 1: Solving Exponential Equations by the Same-Base Method (22 min)

Plain language first. An exponential equation has the variable in the exponent. When you can rewrite both sides as powers of the same base, the exponents must be equal — because the only way two equal powers of the same base can exist is if their exponents match.

The one-line rule: If bᵐ = bⁿ and b > 0, b ≠ 1, then m = n.

Fully worked example 1 — straightforward:

Solve 2ˣ = 16.
1. Ask: is 16 a power of 2? Yes: 16 = 2⁴.
2. Rewrite: 2ˣ = 2⁴.
3. Equate exponents: x = 4.
4. Check: 2⁴ = 16. ✓
The key step is recognizing the base. Students lose points by writing 2x = 4 (doubling instead of equating).

Fully worked example 2 — exponent has a linear expression:

Solve 3^(x+1) = 27.
1. Rewrite 27 as a power of 3: 27 = 3³.
2. So 3^(x+1) = 3³ → x + 1 = 3 → x = 2.
3. Check: 3^(2+1) = 3³ = 27. ✓

Fully worked example 3 — different bases, convert both:

Solve 9ˣ = 3^(x+4).
1. Rewrite 9 = 3²: (3²)ˣ = 3^(x+4) → 3^(2x) = 3^(x+4).
2. Equate: 2x = x + 4 → x = 4.
3. Check: 9⁴ = 6561 and 3^(4+4) = 3⁸ = 6561. ✓

Named misconception + cure:
- ❌ "3^(x+1) = 27 → I multiply 3 · (x+1) = 27 → x = 8." (treating the exponent like a factor)
✅ Cure: an exponent is not a multiplier — 3^(x+1) means 3 multiplied by itself (x+1) times. The equation 3^(x+1) = 3³ is solved by matching exponents (x + 1 = 3), not by multiplying the base by the exponent.

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Segment 3 — Method 2: Taking a Logarithm of Both Sides (25 min)

Bridge. "What happens when the two sides can't be written with the same base? For example, 2ˣ = 10 — 10 is not a power of 2. Same-base method fails. We need another tool: take the log of both sides."

The one-line rule: If aˣ = b and b > 0, then x = log_a(b) = ln(b)/ln(a) (change of base).

Fully worked example 4 — base 2, not a power-of-2 target:

Solve 2ˣ = 10.
1. Take log of both sides (you can use ln or log base 10 — both work; use ln here):
ln(2ˣ) = ln(10)
2. Power rule for logs — bring the exponent down:
x · ln 2 = ln 10
3. Divide both sides by ln 2:
x = ln(10) / ln(2) = log₂(10) ≈ 3.322
4. Check: 2^3.322 ≈ 9.998 ≈ 10. ✓
Exact answer: x = ln 10 / ln 2. Decimal: ≈ 3.32.

Memory hook: "When the exponent is the unknown, use logarithms to get it out of the sky and back on the ground."

Fully worked example 5 — compound interest equation, solving for t:

$1,000 invested at 6% compounded continuously; find the doubling time.
A = Pe^(rt) → 2000 = 1000 · e^(0.06t)
1. Divide both sides by 1000: 2 = e^(0.06t)
2. Take ln of both sides: ln 2 = 0.06t
3. Divide: t = ln(2) / 0.06 ≈ 11.55 years
Exact answer: t = ln 2 / 0.06. Meaning: your money doubles in about 11.6 years at 6% continuous interest.

Named misconception + cure:
- ❌ "I take log of both sides of 5^x = 200 and get 5x = log(200) = 2.301." (trying to pull the base-exponent down as a product)
✅ Cure: the power rule says log(5^x) = x · log(5), NOT 5x. Write it step by step: log(5^x) → x·log 5 = log 200 → x = log(200)/log(5). The base stays in the denominator.

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Segment 4 — Solving Logarithmic Equations + the Extraneous Solution Trap (18 min) · Session 1 closes (~73 min)

Bridge. "Now flip it: what if the variable is inside a logarithm?"

Two-step strategy:
1. Condense — use log properties to rewrite as a single log on one side.
2. Convert — rewrite log_b(expression) = c as b^c = expression; then solve.
3. Check the domain — every log argument must be positive; reject extraneous solutions.

Fully worked example 6 — single log, simple:

Solve log₂(x − 1) = 3.
1. Convert: x − 1 = 2³ = 8 → x = 9.
2. Check: x − 1 = 8 > 0. ✓ And log₂(8) = 3. ✓

Fully worked example 7 — two logs to combine (THE extraneous-solution example):

Solve log(x) + log(x − 3) = 1 [base 10].
1. Condense (product rule): log[x(x − 3)] = 1
2. Convert: x(x − 3) = 10¹ = 10 → x² − 3x − 10 = 0
3. Factor: (x − 5)(x + 2) = 0 → x = 5 or x = −2
4. Domain check:
- x = 5: log(5) and log(5−3) = log(2) — both positive. ✓
- x = −2: log(−2) — negative argument. Extraneous — discard.
5. Answer: x = 5 only.
The trap: x = −2 looks like a valid algebra solution and the quadratic gives it cleanly. But log(−2) is undefined — always check.

Named misconception + cure:
- ❌ "Both x = 5 and x = −2 are answers — I solved the quadratic correctly."
✅ Cure: solving the quadratic correctly gets you the candidates; checking the domain gets you the valid answers. In a log equation, every argument must be strictly positive in the original equation. x = −2 makes log(−2) undefined — it's eliminated by the domain, not the algebra.

Quick interaction — think-pair-share (5 min):
Put on the board: "log₂(x) + log₂(x + 2) = 3 — step 1 is to condense. What does that give you?" Students pair for 30 sec, class votes. Answer: log₂(x(x + 2)) = 3, which becomes x(x+2) = 8, → x² + 2x − 8 = 0, → (x + 4)(x − 2) = 0. Check: x = 2 valid (log₂(2) + log₂(4) = 1 + 2 = 3 ✓); x = −4 extraneous. Answer: x = 2.

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Segment 5 — Applications Part 1: Compound Interest & Solving for Time (25 min) · Session 2 opens

Hook back in. "Everything we've done this week has been prep for this: real-world growth and decay problems where you need the time."

The two main formulas:
- Continuous compounding: A = Pe^(rt). Use when the problem says "continuously."
- Periodic compounding: A = P(1 + r/n)^(nt). Use when the problem says n times per year.

Doubling time — worked example:

How long to triple $2,000 at 5% compounded continuously?
1. Set up: 3(2000) = 2000 · e^(0.05t) → 3 = e^(0.05t)
2. Take ln: ln 3 = 0.05t
3. Solve: t = ln(3)/0.05 ≈ 21.97 years
4. Interpret: at 5% continuous compounding, your money triples in about 22 years.

General doubling-time formula (derive it, don't just hand it):

2P = Pe^(rt) → 2 = e^(rt) → t = ln(2)/r.
At 6%, t = ln(2)/0.06 ≈ 11.55 yr. At 8%, t = ln(2)/0.08 ≈ 8.66 yr. The rule of 72 (divide 72 by the rate %) gives a quick estimate: 72/6 = 12 yr.

Named misconception + cure:
- ❌ "I divide both sides by P to get 2 = e^(rt) and then I write 2/e = rt." (algebraically dividing by e instead of taking ln)
✅ Cure: e is not a factor — e^(rt) is "e to the power rt." You undo a power of e by applying ln to both sides: ln(2) = ln(e^(rt)) = rt. Never divide by e.

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Segment 6 — Applications Part 2: Decay, Half-Life & General Models (18 min)

Half-life model: A(t) = A₀ · e^(−kt), where k > 0 is the decay constant.

The half-life connection: when A = A₀/2, the equation gives e^(−kt₁/₂) = 1/2, so t₁/₂ = ln(2)/k.

Worked example — carbon-14-style decay:

A 400 g sample decays at rate k = 0.15/hr. How long to reach 50 g?
1. Set up: 50 = 400 · e^(−0.15t)
2. Divide: 1/8 = e^(−0.15t)
3. Take ln: ln(1/8) = −0.15t → −3 ln 2 = −0.15t
4. Solve: t = 3 ln(2)/0.15 = ln(8)/0.15 ≈ 13.86 hours
5. Sanity check: each "half-life" ≈ ln(2)/0.15 ≈ 4.62 hr; 400 → 200 → 100 → 50 is 3 half-lives = 13.86 hr. ✓

Named misconception + cure:
- ❌ "The decay rate is −0.15, so I write A = A₀ − 0.15t." (mixing up linear and exponential decay)
✅ Cure: exponential decay multiplies the amount by a factor that shrinks over time — A = A₀ · e^(−kt). Subtracting 0.15t per hour is linear and is wrong here. If the problem says "decays at a continuous rate," the model is exponential.

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Segment 7 — Technology Workflow + AI-Critique Moment (15 min)

Technology workflow — verify a solution graphically in Desmos:
1. Go to desmos.com/calculator.
2. Graph y = 2^x on line 1 and y = 10 on line 2.
3. Find the intersection point (they cross near x ≈ 3.32).
4. Compare to your algebraic answer: x = ln(10)/ln(2) ≈ 3.322. Match → algebra confirmed.
5. Same trick works for log equations: graph y = log(x) + log(x − 3) and y = 1; intersection at x = 5. ✓

AI-critique moment (students verify, not consume):

Paste this to an approved chatbot: "Solve log(x) + log(x + 4) = 1 and find all real solutions."
What to watch for: chatbots frequently accept both algebraic roots and forget to check the domain. The equation gives x² + 4x = 10 → x² + 4x − 10 = 0 → x = (−4 ± √56)/2 = −2 ± √14. x = −2 + √14 ≈ 1.742 (valid: both log arguments positive); x = −2 − √14 ≈ −5.742 (extraneous: log(−5.742) undefined). Your job: check whether the chatbot discards x = −2 − √14 as extraneous. If it doesn't, it's wrong. The tool drafts; you judge.

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Segment 8 — Callback, Tease & Hand-off (12 min) · Session 2 closes (~77 min)

Callback:
- Week 13: exponential functions — what they look like, how they grow or decay.
- Week 14: logarithms as inverses — log properties, change of base.
- Week 15 (this week): solving equations — putting the inverse relationship to work to solve for the unknown in an exponent or inside a log. Every technique here (same-base, log-of-both-sides, condense-then-convert, domain check) comes directly from what you learned in Weeks 13–14.

Tease next week — Week 16:
"Next week is your final exam. It's cumulative over all eight objectives — from exponent rules and linear equations all the way through exponential and logarithmic applications. But look at what you know: you can solve equations in the exponent, you can find 'how long until?' for any growth or decay model, and you know to check for extraneous solutions. Those are the Week 15 items on the final — and you just did them."

Hand-off (the week's graded work):
- Lecture Tutorial 15 (AI tutor, share-link submission) — same-base method, log-of-both-sides, log equations, applications.
- Quiz 15 (end of week, no AI) — 10 items covering all four skill clusters.
- Discussion 15 ("How Long Until…?") — pick a real-world model and set it up.
- Assignment 15 ("Solving Exponential & Logarithmic Equations") — AI-coached, self-scored, four problems.

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Instructor FAQ — Common Stumbles

Student says / does	Quick cure
"2^x = 10, so I write 2x = 10 and x = 5."	The equation is 2 to the power x, not 2 times x. Take log of both sides: x·ln 2 = ln 10, so x = ln10/ln2 ≈ 3.32.
"I take the log of both sides of 5^x = 200 and get 5x = log 200."	Power rule says log(5^x) = x·log 5, NOT 5x. Write it one step at a time: x·log 5 = log 200 → x = log(200)/log(5).
Accepts both x = 5 and x = −2 as solutions to log(x) + log(x−3) = 1.	Domain check is required: log(−2) is undefined. Plug x = −2 into the original equation — it produces a log of a negative number. Extraneous — discard. Only x = 5 is valid.
"A = A₀ − kt for decay" (uses linear instead of exponential model).	Exponential decay means the amount decreases by a factor each period, not by a fixed amount. The model is A = A₀·e^(−kt).
"To undo e^(rt) = 2, I divide by e and get rt = 2/e."	e is a base, not a factor. Take ln of both sides: ln(e^(rt)) = ln 2 → rt = ln 2.
"9^x = 3^(x+4) — I just set 9x = 3(x+4)." (equating the wrong things)	Rewrite 9 = 3² first: (3²)^x = 3^(x+4) → 3^(2x) = 3^(x+4). NOW set the exponents equal: 2x = x + 4.
"I don't know which log to use."	Any base works. For equations with e, use ln (natural log) — it cancels cleanly. For others, use log (base 10) or whichever base matches the equation. The answer is the same regardless.
"ln(2)/0.06 — which key is that on my calculator?"	ln key × 2, then ÷ 0.06. Most scientific calculators have a dedicated ln key. Alternatively, log(2)/log(e) = log(2)/0.4343 ≈ 0.301/0.434 ≈ the same answer.

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Scope flag

This outline stays within Objective 8, focusing on solving exponential and logarithmic equations and the three main application types (compound interest, doubling/tripling time, and exponential decay/half-life). Logistics of general periodic compounding A = P(1 + r/n)^(nt) are set up and mentioned but the main solving work uses continuous compounding e^(rt) for cleaner algebra. Logistic growth models and base-changing for non-standard bases are out of scope for this week.