log_b c = a ⟺ bᵃ = c fluently in both directionsRiddle: 2 raised to what power gives 32?
You just computed a logarithm: log₂32 = 5. That's all a logarithm is — an exponent with a question mark on it. "log base 2 of 32" asks "what exponent turns 2 into 32?"
Every scary-looking log fact is an exponent fact wearing a disguise. Yesterday you built inverse functions; today's entire lesson is: the logarithm is the inverse of the exponential, plus the paperwork.
log_b c = a ⟺ bᵃ = c (b > 0, b ≠ 1, c > 0)
Practice the translation until it's reflexive: - log₂32 = 5 because 2⁵ = 32 - log₃(1/9) = −2 because 3⁻² = 1/9 - log_b 1 = 0 because b⁰ = 1 (any base) - log_b b = 1 because b¹ = b
Special bases: log x = log₁₀x (common log); ln x = log_e x (natural log). Same rules, favorite bases.
f(x) = log_b x is f⁻¹ for bˣ. Lesson 14's swap rules deliver everything:
| exponential bˣ (b > 1) | logarithmic log_b x |
|---|---|
| domain ℝ | domain (0, ∞) |
| range (0, ∞) | range ℝ |
| horizontal asymptote y = 0 | vertical asymptote x = 0 |
| point (0, 1) | point (1, 0) |
| grows faster and faster (concave up) | grows slower and slower (concave down), yet still unbounded |
| always increasing | always increasing |
Inverse compositions, in log clothes:
log_b(bˣ) = x b^(log_b x) = x (x > 0)
End behavior of log_b x (b > 1): lim x→0⁺ log_b x = −∞ (plunges down the asymptote) and lim x→∞ log_b x = ∞ — unbounded, but with no horizontal asymptote and agonizing slowness (log₂ of a billion is only ≈ 30). Like exponentials, log functions have no extrema and no inflection points on their domain.
[GRAPH: y = 2ˣ and y = log₂x reflected across dashed line y = x. Exponential through (0,1) and (1,2) hugging y = 0 leftward; log through (1,0) and (2,1) hugging x = 0 downward. Asymptotes dashed and labeled.]
Transformations apply as always: log₅(x − 4) shifts right 4 → domain x > 4, asymptote x = 4.
Each log property is an exponent property read through the inverse:
Product: log_b(MN) = log_b M + log_b N (bᵐ·bⁿ = bᵐ⁺ⁿ)
Quotient: log_b(M/N) = log_b M − log_b N (bᵐ/bⁿ = bᵐ⁻ⁿ)
Power: log_b(Mᵖ) = p·log_b M ((bᵐ)ᵖ = bᵐᵖ)
Expand (one log → sum/differences) and condense (many logs → one) are the two directions of FRQ 3's favorite task:
Expand: log(a b²) = log a + 2 log b
Condense: 3 ln x − ln y = ln(x³) − ln y = ln(x³/y)
What the properties do NOT say (the forbidden moves): - log(M + N) ≠ log M + log N — no property touches a sum inside - (log M)(log N) ≠ log(MN) — the product property adds logs, it doesn't multiply them - log M / log N ≠ log(M/N) — that quotient is actually a change-of-base expression!
log_b c = log_a c / log_a b usually: log_b c = ln c / ln b
This is how you evaluate log₇20 on any calculator, and how all log functions reveal themselves as vertical scalings of one another: log₇x = (1/ln 7)·ln x. One log curve, many rulers.
Problem: Evaluate: (a) log₄64 (b) log₉3 (c) log₅(1/25) (d) ln(e⁷)
Solution: (a) 4^? = 64 → 3 (b) 9^? = 3 → 9^(1/2) = 3 → 1/2 (c) 5^? = 1/25 → −2 (d) inverse composition → 7
Interpretation: Say the sentence "base to what power gives the input?" — fractions mean roots, reciprocals mean negatives.
Problem: For g(x) = 2·log₃(x + 6) − 1: domain, asymptote, x-intercept behavior, end behavior.
Solution: Need x + 6 > 0: domain x > −6; vertical asymptote x = −6. End behavior: lim x→−6⁺ g(x) = −∞, lim x→∞ g(x) = ∞. Increasing and concave down throughout. x-intercept: solve 2log₃(x+6) = 1 → log₃(x+6) = 1/2 → x + 6 = 3^(1/2) → x = √3 − 6 ≈ −4.27.
Interpretation: The inside expression drives the domain and asymptote; the outside constants (2, −1) only stretch and slide vertically — they never move the asymptote's location.
Problem: (a) Expand fully: ln( x³√y / z² ) (with x, y, z > 0). (b) Condense: 2 log(x) + (1/2)log(y) − 3 log(z).
Solution: (a)
ln(x³ y^(1/2) / z²) = 3 ln x + (1/2) ln y − 2 ln z
(b)
log(x²) + log(y^(1/2)) − log(z³) = log( x²√y / z³ )
Interpretation: Power property first when expanding (pull exponents down), last when condensing (push coefficients up). Multiplication ↔ addition, division ↔ subtraction, exponent ↔ coefficient — the full dictionary.
Problem: Express log₇20 in terms of natural logs, and determine between which two consecutive integers it lies without a calculator.
Solution: log₇20 = ln 20 / ln 7. Bracket with powers of 7: 7¹ = 7 < 20 < 49 = 7², so 1 < log₇20 < 2.
Interpretation: The bracketing trick — squeeze the input between two powers of the base — answers every "between which integers" log question in seconds and sanity-checks any calculator value.
1. (A). 2⁵ = 32 → 5. (C) is log₂16; (B) answers "half of 32."
2. (B). 3⁻² = 1/9 → −2. (A) drops the sign; reciprocal inputs give negative logs.
3. (C). Inside > 0: x − 4 > 0 → x > 4 (strict — the asymptote itself is excluded, so (B) fails).
4. (D). Inside zero at x = −3 → vertical asymptote x = −3. (B) flips the shift; (A)/(C) confuse vertical with horizontal.
5. (A). Product then power: log a + log(b²) = log a + 2 log b. (B) squares the whole product; (C)/(D) multiply logs — the forbidden move.
6. (B). ln(x³) − ln y = ln(x³/y). (A) treats the coefficient 3 as a factor inside; (C) subtracts inside (forbidden); (D) applies the 3 to both.
7. (A). Change of base, input on top: ln 20/ln 7 ≈ 1.5 ✓ (sanity: between 1 and 2 since 7 < 20 < 49). (B) is upside down (≈ 0.65 — too small).
8. (C). log_b 1 = 0 always → (1, 0). (A) belongs to the exponential; (B) is on neither (log undefined at 0).
9. (B). Inverse of base-3 exponential is log₃x (the swap: 3ˣ answers "3 to the x"; log₃ asks "3 to the what?"). (A) inverts x³; (C) reflects, not inverts.
10. (D). "b to what power gives b⁷?" → 7. Inverse composition — no computation.
lim x→0⁺ ln x =11. (B). The log plunges down its asymptote: −∞. (A) confuses with ln 1 = 0.
12. (FRQ-style, 6 points)
(i) [2 pts] Domain x > 1; range ℝ; vertical asymptote x = 1 with lim x→1⁺ g(x) = −∞; also lim x→∞ g(x) = ∞.
(ii) [2 pts] y = log₂(x − 1) + 3 → log₂(x − 1) = y − 3 → x = 2^(y−3) + 1: g⁻¹(x) = 2^(x−3) + 1. Verify: g(g⁻¹(5)) = g(2² + 1) = g(5) = log₂4 + 3 = 2 + 3 = 5 ✓
(iii) [2 pts] Shift y = log₂x right 1, then up 3. (Order: horizontal shift, then vertical shift; the asymptote moves with the horizontal shift, from x = 0 to x = 1.)
12. (FRQ-style) 🚫 Let g(x) = log₂(x − 1) + 3. (i) State the domain, range, and vertical asymptote of g, with limit notation for behavior near the asymptote. (ii) Find g⁻¹(x) and verify that g(g⁻¹(5)) = 5. (iii) Describe the graph of g as transformations of y = log₂x.
(a) (i) Rewrite as a single logarithm: 2 log₆x + log₆(x − 5) − log₆4 (x > 5).
(ii) Evaluate log₆9 + log₆4 exactly.
(b) Given only that ln 2 ≈ 0.69 and ln 3 ≈ 1.10, evaluate each without a calculator: (i) ln 12 (ii) ln(9/2) (iii) log₃8.
(c) The function f(x) = 5·bˣ (b > 1) has inverse f⁻¹. (i) Find a formula for f⁻¹(x) using logarithms base b. (ii) Verify f⁻¹(5b²) = 2.
(a) [2 pts] (i) [1 pt] log₆(x²) + log₆(x − 5) − log₆4 = log₆( x²(x − 5)/4 ). (ii) [1 pt] log₆(9·4) = log₆36 = 2.
(b) [2 pts] (i) ln 12 = ln(4·3) = 2 ln 2 + ln 3 ≈ 1.38 + 1.10 = 2.48 [1 pt] (ii) ln(9/2) = 2 ln 3 − ln 2 ≈ 2.20 − 0.69 = 1.51; (iii) log₃8 = ln 8/ln 3 = 3 ln 2/ln 3 ≈ 2.07/1.10 ≈ 1.88 [1 pt for both, using properties + change of base]
(c) [2 pts] (i) [1 pt] y = 5bˣ → bˣ = y/5 → x = log_b(y/5): f⁻¹(x) = log_b(x/5). (ii) [1 pt] f⁻¹(5b²) = log_b(5b²/5) = log_b(b²) = 2 ✓
1. (A). 2⁵ = 32 → 5. (C) is log₂16; (B) answers "half of 32."
2. (B). 3⁻² = 1/9 → −2. (A) drops the sign; reciprocal inputs give negative logs.
3. (C). Inside > 0: x − 4 > 0 → x > 4 (strict — the asymptote itself is excluded, so (B) fails).
4. (D). Inside zero at x = −3 → vertical asymptote x = −3. (B) flips the shift; (A)/(C) confuse vertical with horizontal.
5. (A). Product then power: log a + log(b²) = log a + 2 log b. (B) squares the whole product; (C)/(D) multiply logs — the forbidden move.
6. (B). ln(x³) − ln y = ln(x³/y). (A) treats the coefficient 3 as a factor inside; (C) subtracts inside (forbidden); (D) applies the 3 to both.
7. (A). Change of base, input on top: ln 20/ln 7 ≈ 1.5 ✓ (sanity: between 1 and 2 since 7 < 20 < 49). (B) is upside down (≈ 0.65 — too small).
8. (C). log_b 1 = 0 always → (1, 0). (A) belongs to the exponential; (B) is on neither (log undefined at 0).
9. (B). Inverse of base-3 exponential is log₃x (the swap: 3ˣ answers "3 to the x"; log₃ asks "3 to the what?"). (A) inverts x³; (C) reflects, not inverts.
10. (D). "b to what power gives b⁷?" → 7. Inverse composition — no computation.
11. (B). The log plunges down its asymptote: −∞. (A) confuses with ln 1 = 0.
12. (FRQ-style, 6 points)
(i) [2 pts] Domain x > 1; range ℝ; vertical asymptote x = 1 with lim x→1⁺ g(x) = −∞; also lim x→∞ g(x) = ∞.
(ii) [2 pts] y = log₂(x − 1) + 3 → log₂(x − 1) = y − 3 → x = 2^(y−3) + 1: g⁻¹(x) = 2^(x−3) + 1. Verify: g(g⁻¹(5)) = g(2² + 1) = g(5) = log₂4 + 3 = 2 + 3 = 5 ✓
(iii) [2 pts] Shift y = log₂x right 1, then up 3. (Order: horizontal shift, then vertical shift; the asymptote moves with the horizontal shift, from x = 0 to x = 1.)
🎯 Exam tip: Before manipulating any log expression, whisper the two forbidden moves: no splitting sums inside, no multiplying logs together. Then use the dictionary — multiply↔add, divide↔subtract, power↔coefficient — one operation at a time. FRQ 3 rewards slow, legal steps over fast, illegal ones.