Week 14 — Lecture Outline · Logarithmic Functions

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objectives covered: Objective 8 — Analyze and apply exponential and logarithmic functions and equations.
SLOs touched: A (apply procedures accurately) · B (connect symbolic/numerical/graphical representations)
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	"A logarithm is just an exponent in disguise — so why does that matter?"
By the end of the week, students can…	(1) define a logarithm as the inverse of an exponential and convert between log and exponential form; (2) evaluate common (base 10) and natural (base e) logarithms, as well as logs in other bases, mentally; (3) identify the domain and vertical asymptote of a logarithmic function; (4) apply the three properties of logarithms — product, quotient, and power — to expand and condense expressions.
Key vocabulary	logarithm, base, argument, inverse function, common logarithm (log), natural logarithm (ln), domain, vertical asymptote, product rule, quotient rule, power rule, expand, condense
Materials	slides (Deck 14), the week's readings + video links, Desmos (or GeoGebra), a calculator, one approved chatbot (Gemini / Claude / ChatGPT)
Timing note	8 segments, ~150 min total. Session 1 = Segments 1–4 (~75 min). Session 2 = Segments 5–8 (~75 min).

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

Hook. "Raise your hand if you've ever heard of a 7.0 earthquake versus a 9.0 earthquake. Most people guess the 9.0 is about twice as strong. It's actually 100 times more energy. Why? Because the Richter scale isn't counting — it's logging."

• "pH 2 (stomach acid) vs. pH 7 (water) — that's a factor of 100,000 in hydrogen-ion concentration, disguised in a 5-point difference. Decibels, star brightness, musical pitch — the same compression trick. And the tool doing all of it is the logarithm."
• "A logarithm is not a mystery. It's the answer to one question: '2 to what power gives you 8?' The answer is 3. That's log₂(8). We just gave the question a name."

The promise (write it on the board): "By the end of this week you can read a log expression out loud in plain English, find the domain of any log function, and expand or condense any log expression using three clean rules — the same three rules that make earthquake-magnitude comparisons possible."

Why it matters: "Every time you read a scientific graph on a log scale, check a pH, or compare sound levels, you're reading a logarithm. This week you'll understand what you're reading — not just use it."

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Segment 2 — The Logarithm as the Inverse of an Exponential (22 min)

Plain language first. A logarithm is the question the exponential answers.
- The exponential 2³ = 8 says: "2 to the power 3 is 8."
- The logarithm log₂(8) = 3 asks: "What power of 2 gives 8?" and answers: 3.
- They're inverses: the exponential takes the exponent and outputs the value; the logarithm takes the value and outputs the exponent.

The formal definition (the "magic swap"):

log_b(x) = y ⟺ bʸ = x, where b > 0, b ≠ 1, and x > 0.
"The log is the exponent you need."

Write the swap on the board:

Log form	↔	Exponential form
log_b(x) = y		bʸ = x
log₂(8) = 3		2³ = 8
log₃(9) = 2		3² = 9
log₁₀(1000) = 3		10³ = 1000
ln(e⁴) = 4		e⁴ = e⁴

One fully worked example — converting both directions:

(A) Rewrite log₄(64) = 3 in exponential form.
Log form tells us: base = 4, exponent = 3, result = 64.
Exponential form: 4³ = 64. (Check: 4³ = 4·4·4 = 64 ✓)

(B) Rewrite 5² = 25 in log form.
Exponential: base = 5, exponent = 2, result = 25.
Log form: log₅(25) = 2. (Check: 5² = 25 ✓)

The common and natural logs (two special cases):
- Common log: log(x) means log₁₀(x) — base 10. This is the one on your calculator's [LOG] key.
- Natural log: ln(x) means logₑ(x) — base e ≈ 2.71828. This is the [LN] key. The e here is the same natural base from Week 13's continuous growth.

Memory hook: "log without a base means base 10; ln means base e — the two special bases that every calculator knows."

Evaluating logs mentally — the one question:

"What power of the base gives the argument?"
- log₂(8): 2 to what power = 8? Since 2³ = 8, log₂(8) = 3.
- log₅(1): 5 to what power = 1? Since 5⁰ = 1, log₅(1) = 0. (Any log of 1 is 0.)
- ln(e): e to what power = e? Since e¹ = e, ln(e) = 1. (Any log of its own base is 1.)
- log₁₀(1000): 10 to what power = 1000? Since 10³ = 1000, log₁₀(1000) = 3.
- log₂(1/4): 2 to what power = 1/4? Since 2⁻² = 1/4, log₂(1/4) = −2. (Reciprocal → negative exponent.)

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Segment 3 — Domain and Graph Features of Logarithmic Functions (20 min)

Plain language first. The domain of a log function has ONE rule: the argument must be strictly positive. You cannot take the log of zero or a negative number.

Why? Because there's no real power of a positive base that gives you 0 or a negative number. (5ˣ is always positive for real x.) So log(negative) and log(0) simply don't exist in the real-number world.

Domain: set argument > 0 and solve.

Example: Find the domain of f(x) = log₃(x − 2).
Set the argument > 0: x − 2 > 0 → x > 2. Domain: (2, ∞).
The vertical asymptote (VA) is where the argument equals zero: x − 2 = 0 → x = 2. The graph approaches x = 2 but never crosses.

Graph features of f(x) = log_b(x) (b > 1):
- Domain: (0, ∞) — only positive x-values.
- Range: (−∞, ∞) — the log output can be any real number.
- Vertical asymptote: x = 0 (the y-axis).
- x-intercept at (1, 0) (because log_b(1) = 0 for any base b).
- Passes through (b, 1) (because log_b(b) = 1).
- Increasing (for b > 1): bigger argument → bigger log value.

Desmos check (exact steps):
1. Open desmos.com/calculator.
2. Type y = log(x - 2) (Canvas log = base 10).
3. Observe the vertical asymptote at x = 2 and domain (2, ∞).
4. Zoom in near x = 2 to watch the graph dive toward −∞.

Two worked examples:

(A) f(x) = log(x + 5). Domain? VA?
Argument: x + 5 > 0 → x > −5. Domain: (−5, ∞).
VA: x + 5 = 0 → x = −5.

(B) f(x) = ln(2x − 1). Domain? VA?
Argument: 2x − 1 > 0 → 2x > 1 → x > 1/2. Domain: (1/2, ∞).
VA: 2x − 1 = 0 → x = 1/2.

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Segment 4 — Misconceptions + Quick Interaction (25 min) · Session 1 closes (~75)

Name the misconceptions out loud, then cure each:

• ❌ "log(M + N) = log M + log N."
  ✅ Cure: WRONG. The product rule applies to multiplication inside the argument, not addition. log(3 + 4) = log(7) ≈ 0.845, but log(3) + log(4) = log(12) ≈ 1.079. Different numbers — the rule doesn't apply. There is NO log property for a sum or difference inside the argument.

• ❌ "I can take log(−5) — it's just negative."
  ✅ Cure: Undefined. The argument must be strictly positive. log(−5) has no real-number value. If a domain problem produces a negative argument, that x-value is NOT in the domain.

• ❌ "log₂(8) = 8/2 = 4." (dividing the argument by the base)
  ✅ Cure: A log is an exponent, not a division. log₂(8) asks "2 to what power gives 8?" The answer is 3, because 2³ = 8. Never divide.

• ❌ "The base and the argument are the same thing."
  ✅ Cure: In log_b(x), b is the base (what you're raising to a power) and x is the argument (the result). log₂(8): base = 2, argument = 8. The answer is the exponent (3). Three separate roles.

Interaction — Quick-Fire Evaluation (Think-Pair-Share, ~10 min):
Post five log expressions on a slide; students evaluate solo (30 sec), compare with neighbor (1 min), class shares:
1. log₃(27) → 3 (3³ = 27)
2. log₁₀(100) → 2 (10² = 100)
3. log₂(1) → 0 (2⁰ = 1)
4. ln(e²) → 2 (eˣ = e² → x = 2)
5. log₅(1/25) → −2 (5⁻² = 1/25)

Debrief #3 and #5 carefully — log of 1 always = 0, negative results always come from reciprocal arguments.

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Segment 5 — The Three Properties of Logarithms (25 min) · Session 2 opens

Hook back in: "Last session: the log is an exponent in disguise. Today: the rules for the algebra of logarithms — and they mirror the exponent rules almost perfectly, because logs are exponents."

Plain language first — three properties, three rules:

Property	Rule	Plain language
Product	log_b(MN) = log_b M + log_b N	Log of a product → sum of logs
Quotient	log_b(M/N) = log_b M − log_b N	Log of a quotient → difference of logs
Power	log_b(Mᵖ) = p · log_b M	Exponent inside → multiplier outside

Why they work: Because logs are exponents, and exponent rules say: multiply same base → add exponents (product rule), divide same base → subtract exponents (quotient rule), power of a power → multiply exponents (power rule). The log properties are just those rules in disguise.

Two key special values (derived from the definition, not from the three rules):
- log_b(b) = 1 (because b¹ = b)
- log_b(1) = 0 (because b⁰ = 1)

Fully worked example — EXPANDING:

Expand log_b(x³y²).
1. Product rule: log_b(x³) + log_b(y²)
2. Power rule: 3·log_b(x) + 2·log_b(y)
(Each factor inside the log separates; each exponent moves outside.)

Fully worked example — CONDENSING:

Condense 2·log(x) + log(y).
1. Power rule (reverse): log(x²) + log(y)
2. Product rule (reverse): log(x²y)
(Move the multiplier inside as an exponent first, then combine the sum into one log.)

One more worked example — EXPANDING with quotient and power:

Expand ln(x⁴/y³).
1. Quotient rule: ln(x⁴) − ln(y³)
2. Power rule on each term: 4·ln(x) − 3·ln(y)

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Segment 6 — Applying the Properties + the Famous Trap (18 min)

The signature trap — the most missed item on every log test:

❌ "log(M + N) = log M + log N."
✅ The product rule says log(MN) — multiplication inside separates into a sum. Addition inside? There is NO rule. You cannot simplify log(x + 5). Full stop.

Show the numerical proof:
- log(3 + 4) = log(7) ≈ 0.845
- log(3) + log(4) = log(12) ≈ 1.079
- These are different numbers — the "rule" is false.

Condensing practice (important for Week 15 equation-solving):

Condense log_b(x) − log_b(y).
Quotient rule: log_b(x/y). Done.

Condense 3·log x − log y.
1. Power rule: log(x³) − log(y)
2. Quotient rule: log(x³/y)

Expanding practice:

Expand log_b(x²·z³).
1. Product rule: log_b(x²) + log_b(z³)
2. Power rule: 2·log_b(x) + 3·log_b(z)

Callback to Week 13: The natural log was introduced alongside e. Now we see: ln(e^k) = k every time, because ln just asks "e to what power gives this?" — and the answer is always the exponent. This is what we'll use in Week 15 to solve exponential equations.

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

Technology workflow — check a log property in Desmos (exact steps):
1. Open desmos.com/calculator.
2. Type the original expression on line 1: log(x^3 * y^2) — Desmos uses log for base 10.
3. Type the expanded expression on line 2: 3*log(x) + 2*log(y).
4. Check: for any positive x, y, both expressions produce the same value. (In slider mode, watch them move together.)
5. On line 3, type log(x + 5) and on line 4 log(x) + log(5). Watch: different curves — the product rule does NOT apply to addition.

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

Paste this to an approved chatbot: "Expand log(x + 4) using log properties."
Check its response. A chatbot that writes "log(x) + log(4)" is applying the product rule to an addition — which is WRONG. The correct answer is: log(x + 4) cannot be simplified using log properties (there's no rule for a sum inside a log). Your job every week: the tool drafts, you judge.

Second AI-critique scenario:

Ask the chatbot: "What is log₂(−8)?"
Any numeric answer is wrong. The correct answer is: undefined (argument must be > 0). If the chatbot gives a real-number answer, it has made an error.

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Segment 8 — Callback, Tease & Hand-off (7 min) · Session 2 closes (~75)

Callback to Week 13:
- Week 13: exponential functions f(x) = bˣ — output is the "power," input is the exponent.
- Week 14: logarithmic functions f(x) = log_b(x) — the exact inverse: output is the exponent, input is the "power."
- Together, they're the most important inverse pair in the course.

Tease Week 15:
- "Now that you understand the logarithm, we'll use it. Week 15: exponential and logarithmic equations — solving things like 2ˣ = 50 (take the log of both sides) and log(x + 3) = 2 (convert to exponential form). The tools are all in your hands."

Hand-off (the week's graded work):
- Lecture Tutorial 14 (AI tutor, share-link submission).
- Quiz 14 (end of week, no AI).
- Discussion 14 — logarithmic scales: why are they useful?
- Assignment 14 — "The Logarithm: An Exponent in Disguise" (AI-coached, self-scored).

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Instructor FAQ — Common Stumbles

Student says / does	Quick cure
"log(x + 5) = log(x) + log(5)"	NO log property for sums inside the argument. Only multiplication separates: log(xy) = log x + log y.
Writes log₂(8) = 4 (divides 8 by 2)	A log is an exponent, not a ratio. Ask: "2 to what power gives 8?" 2³ = 8, so log₂(8) = 3.
Writes domain of log(x − 3) as x ≥ 3	Strict inequality: argument > 0, so x − 3 > 0 means x > 3 (not ≥). The endpoint x = 3 makes the argument 0, which is undefined.
Takes log(−5) and gets −0.699 (using the calculator)	Calculator is in complex mode. log(negative) is undefined in the reals. Domain check first.
"log_b(M/N) = log_b M · log_b N" (multiplied instead of subtracted)	Quotient rule: subtraction outside, division inside. log_b(M/N) = log_b M − log_b N.
Confuses base and argument in log₂(8)	base = 2 (bottom), argument = 8 (inside). The log = 3 is the exponent. Three roles: base, argument, exponent.
"ln(e²) = e² or 2e"	ln undoes e: ln(eˣ) = x. So ln(e²) = 2. The exponent comes straight out.
Can't condense 3·log(x) − log(y)	Power rule first: log(x³) − log(y). Then quotient rule: log(x³/y). Always power rule before product/quotient when condensing.

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Scope flag

This outline stays within Objective 8 for the log half of the unit. Change-of-base is mentioned in the readings but is not a tested skill this week (it appears if needed in Week 15 equation-solving). Complex-number solutions to log equations are outside scope.