Exponential and log items span three exam categories at once: Representations (30%), Functions: Concept and Properties (15%), and Functions as Models (10%) — the official sample set even includes a numeric-entry continuous-compounding problem. Section 1's online graphing calculator can evaluate any log and locate intersections; Section 2 expects exact-form answers like ln 9/ln 4 by hand.
f(x) = a·bˣ a ≠ 0, b > 0, b ≠ 1
The number e ≈ 2.718 (between 2 and 3) is the natural base arising from continuous growth; eˣ is an ordinary exponential with base e > 1, so everything above applies. Continuous-compounding models A(t) = A₀e^(rt) use it — a CLEP favorite.
Finding a and b from two points: a = f(0) when x = 0 is given; otherwise divide outputs so a cancels. Through (0, 6) and (2, 54): a = 6, then 6b² = 54 → b² = 9 → b = 3 (bases are positive by definition — discard −3 automatically).
[GRAPH: Two curves in the xy-plane. f(x) = 6·3ˣ rising steeply through (0, 6) and (1, 18), hugging the dashed asymptote y = 0 on the far left. g(x) = 5·(0.7)ˣ falling through (0, 5), flattening toward y = 0 on the far right. Annotation: "b > 1 grows; 0 < b < 1 decays; neither graph ever touches y = 0".]
log_b c = a ⟺ bᵃ = c (b > 0, b ≠ 1, c > 0)
"log base 2 of 32" asks: 2 to what power gives 32? Answer 5. Drill the translation both directions:
y = log_b x is the inverse of y = bˣ, so every feature swaps roles:
| exponential bˣ (b > 1) | logarithmic log_b x |
|---|---|
| domain: all reals | domain: x > 0 |
| range: y > 0 | range: all reals |
| horizontal asymptote y = 0 | vertical asymptote x = 0 |
| passes through (0, 1) | passes through (1, 0) |
| rises faster and faster | rises ever more slowly — but without bound |
Near its asymptote the log plunges downward; to the right it keeps rising forever, just slowly (log₂ of a billion is only about 30). Shifts move the domain with the asymptote: log₅(x − 4) requires x − 4 > 0 — domain x > 4, asymptote x = 4, strict inequality because the asymptote line itself is excluded.
Bracketing trick: to locate log₃50 without a calculator, squeeze 50 between powers of 3: 27 = 3³ < 50 < 3⁴ = 81, so log₃50 is between 3 and 4.
Each log property mirrors an exponent rule:
Product: log_b(MN) = log_b M + log_b N
Quotient: log_b(M/N) = log_b M − log_b N
Power: log_b(Mᵖ) = p·log_b M
Change of base: log_b c = ln c / ln b (the INPUT rides on top)
Forbidden moves the exam plants in distractors: log(M + N) never splits into log M + log N; (log M)(log N) is not log(MN); and log M / log N is not log(M/N) — that quotient is a change-of-base expression.
Route 1 — match the bases: 3ˣ = 81 = 3⁴ → x = 4. Try this first; powers of 2, 3, and 5 cover most no-calculator items. Route 2 — take a log of both sides: 4ˣ = 9 → x = log₄9 = ln 9/ln 4 (exact form).
Isolate the exponential before logging. For 5·2ˣ = 40: divide by 5 first (2ˣ = 8 → x = 3). Two errors to avoid: subtracting the coefficient instead of dividing, and treating the coefficient as the base.
Route 1 — rewrite as an exponential: log₂(x − 3) = 4 → x − 3 = 2⁴ → x = 19. Route 2 — condense first, then exponentiate:
log x + log(x − 3) = 1
log[x(x − 3)] = 1 → x² − 3x = 10 → (x − 5)(x + 2) = 0 → x = 5 or x = −2
Check both in the original: x = 5 gives log 5 + log 2 = log 10 = 1 ✓; x = −2 makes log x undefined ✗. Solution: x = 5 only. Condensing enlarges the domain, so the condensed equation can produce ghosts — the exam always lists the ghost among the answer choices. Checking candidates in the original equation is not optional.
Q1. B. The parent 4·3ˣ levels off at y = 0 on the far left; subtracting 2 shifts the whole graph, asymptote included, down to y = −2. A) forgets the vertical shift. C) grabs the coefficient a = 4, which sets the y-intercept's distance from the asymptote, not the asymptote. D) grabs the base. E) drops the sign of the shift. Fix: in a·bˣ + d, the horizontal asymptote is y = d — the added constant, sign and all.
Q2. E. a = f(0) = 6, so 6b² = 54 → b² = 9 → b = 3, taking the positive root because exponential bases satisfy b > 0. A) reports b² instead of b. B) subtracts the outputs (54 − 6) instead of dividing. C) takes the negative square root, which is not a legal base. D) reports a instead of b. Fix: the y-intercept hands you a; a second point gives bⁿ, and only the positive root survives.
Q3. B. The question "3 to what power gives 1/9?" has answer −2, since 3⁻² = 1/9. A) drops the sign — reciprocal inputs force negative exponents. C) answers "3 to what power gives √3?", confusing reciprocals with roots. D) combines both errors. E) treats the log as division by the base twice, landing on −9. Fix: translate every log to its exponential sentence — fractions of the base mean negative exponents, roots mean fractional ones.
Q4. D. The inside of a logarithm must be strictly positive: x − 4 > 0, so x > 4. A) applies the parent function's domain without the shift. B) includes x = 4, where the inside is 0 and the log is undefined — the asymptote itself is excluded, so the inequality is strict. C) treats the log like a polynomial. E) flips the shift direction. Fix: set inside > 0 (strict) and solve; the vertical asymptote sits exactly where the inside hits zero.
Q5. D. log_b 1 = 0 for every base because b⁰ = 1, so (1, 0) is on every log graph. A) is the exponential's anchor point (b⁰ = 1 read the other way); it lies on y = bˣ, not the log. B) is on neither — the log is undefined at 0. C) reverses coordinates; the point (b, 1) is on the graph, not (b, 0). E) fails since log_b 1 = 0 ≠ 1. Fix: the log graph anchors at (1, 0) and (b, 1) — the exponential's anchors with coordinates swapped.
Q6. B. Power property first: 3 ln x = ln(x³); then the quotient property: ln(x³) − ln y = ln(x³/y). A) treats the coefficient 3 as a factor inside the log instead of an exponent. C) subtracts inside the log — but subtraction of logs means division inside, never subtraction inside. D) applies the 3 to both x and y. E) turns subtraction into multiplication inside. Fix: coefficient ↔ exponent, subtraction ↔ division — apply the dictionary one operation at a time.
Q7. A. Isolate the exponential first: 2ˣ = 40/5 = 8 = 2³, so x = 3. B) solves for 2ˣ correctly but reports the value of the power, 8, instead of the exponent. C) misreads 2ˣ as 2x and solves 10x = 40. D) subtracts the 5 instead of dividing, solving 2ˣ = 35. E) treats the coefficient 5 as the base. Fix: strip the coefficient by division BEFORE undoing the exponent, and answer with the exponent, not the power.
Q8. E. With a = 5 > 0 and base 0.7 < 1, every unit step multiplies the output by 0.7 — the function decreases everywhere. A), C), and D) have bases greater than 1 (3, 4, and e ≈ 2.718) with positive coefficients: all increasing. B) is the trap: base 0.7 suggests decay, but the negative coefficient reflects the graph below the axis, and a curve rising from large-negative values toward 0 is increasing. Fix: monotonicity needs both signs — a > 0 with b < 1 decreases; flipping the sign of a flips the direction.
Q9. C. Change of base: log₇20 = ln 20/ln 7, input on top. Sanity check: 7 < 20 < 49 means the answer is between 1 and 2, and ln 20/ln 7 ≈ 3.0/1.9 ≈ 1.5 ✓. A) inverts the quotient (≈ 0.65 — visibly less than 1, impossible). B) commits the forbidden move log(M)/log(N) = log(M/N) in reverse. D) turns the quotient of logs into a difference inside. E) drops the logs entirely. Fix: log_b c = (log of the input)/(log of the base) — when in doubt, size-check against integer powers of the base.
Q10. A. Condensing enlarged the domain: the product x(x − 3) is positive at x = −2, but the original equation needs each log defined, and log(−2) does not exist — so −2 is extraneous and only x = 5 survives (log 5 + log 2 = log 10 = 1 ✓). B) validates against the condensed equation, which is exactly where ghosts hide. C) rejects a value that checks perfectly. D) has sign-flipped roots of the wrong quadratic. E) quadratics guarantee at most two candidates, not two valid solutions. Fix: after solving any log equation, substitute every candidate into the ORIGINAL equation and discard any that makes an inside nonpositive.
Q11. C. Bracket 50 between powers of 3: 3³ = 27 < 50 < 81 = 3⁴, so 3 < log₃50 < 4. A) brackets with 3¹ = 3 and 3² = 9, stopping before reaching 50. B) checks 3³ = 27 < 50 but never tests the next power, assuming the log stays below 3. D) overshoots by comparing against 3⁴ = 81 and reading the inequality backwards. E) estimates by division (50/9 ≈ 5.6) rather than by exponents. Fix: list powers of the base until they straddle the input — the two exponents that trap it are the answer.
Q12. C. The unknown t is trapped in the exponent; dividing by 0.05 cannot free it. Take the natural log of both sides: 0.05t = ln 2, so t = (ln 2)/0.05 ≈ 13.9 years. A) treats e^(0.05t) as if it were 0.05t — dividing works only on linear equations. B) is a false verification: e^(0.05·40) = e² ≈ 7.39, not 2. D) applies e instead of ln, compounding the confusion. E) uses the common log where the base is e; log₁₀2 ≈ 0.301 gives a wrong answer that looks plausible. Fix: when the unknown is in an exponent with base e, apply ln to both sides — ln e^(u) = u — then divide.
Q1. B. The parent 4·3ˣ levels off at y = 0 on the far left; subtracting 2 shifts the whole graph, asymptote included, down to y = −2. A) forgets the vertical shift. C) grabs the coefficient a = 4, which sets the y-intercept's distance from the asymptote, not the asymptote. D) grabs the base. E) drops the sign of the shift. Fix: in a·bˣ + d, the horizontal asymptote is y = d — the added constant, sign and all.
Q2. E. a = f(0) = 6, so 6b² = 54 → b² = 9 → b = 3, taking the positive root because exponential bases satisfy b > 0. A) reports b² instead of b. B) subtracts the outputs (54 − 6) instead of dividing. C) takes the negative square root, which is not a legal base. D) reports a instead of b. Fix: the y-intercept hands you a; a second point gives bⁿ, and only the positive root survives.
Q3. B. The question "3 to what power gives 1/9?" has answer −2, since 3⁻² = 1/9. A) drops the sign — reciprocal inputs force negative exponents. C) answers "3 to what power gives √3?", confusing reciprocals with roots. D) combines both errors. E) treats the log as division by the base twice, landing on −9. Fix: translate every log to its exponential sentence — fractions of the base mean negative exponents, roots mean fractional ones.
Q4. D. The inside of a logarithm must be strictly positive: x − 4 > 0, so x > 4. A) applies the parent function's domain without the shift. B) includes x = 4, where the inside is 0 and the log is undefined — the asymptote itself is excluded, so the inequality is strict. C) treats the log like a polynomial. E) flips the shift direction. Fix: set inside > 0 (strict) and solve; the vertical asymptote sits exactly where the inside hits zero.
Q5. D. log_b 1 = 0 for every base because b⁰ = 1, so (1, 0) is on every log graph. A) is the exponential's anchor point (b⁰ = 1 read the other way); it lies on y = bˣ, not the log. B) is on neither — the log is undefined at 0. C) reverses coordinates; the point (b, 1) is on the graph, not (b, 0). E) fails since log_b 1 = 0 ≠ 1. Fix: the log graph anchors at (1, 0) and (b, 1) — the exponential's anchors with coordinates swapped.
Q6. B. Power property first: 3 ln x = ln(x³); then the quotient property: ln(x³) − ln y = ln(x³/y). A) treats the coefficient 3 as a factor inside the log instead of an exponent. C) subtracts inside the log — but subtraction of logs means division inside, never subtraction inside. D) applies the 3 to both x and y. E) turns subtraction into multiplication inside. Fix: coefficient ↔ exponent, subtraction ↔ division — apply the dictionary one operation at a time.
Q7. A. Isolate the exponential first: 2ˣ = 40/5 = 8 = 2³, so x = 3. B) solves for 2ˣ correctly but reports the value of the power, 8, instead of the exponent. C) misreads 2ˣ as 2x and solves 10x = 40. D) subtracts the 5 instead of dividing, solving 2ˣ = 35. E) treats the coefficient 5 as the base. Fix: strip the coefficient by division BEFORE undoing the exponent, and answer with the exponent, not the power.
Q8. E. With a = 5 > 0 and base 0.7 < 1, every unit step multiplies the output by 0.7 — the function decreases everywhere. A), C), and D) have bases greater than 1 (3, 4, and e ≈ 2.718) with positive coefficients: all increasing. B) is the trap: base 0.7 suggests decay, but the negative coefficient reflects the graph below the axis, and a curve rising from large-negative values toward 0 is increasing. Fix: monotonicity needs both signs — a > 0 with b < 1 decreases; flipping the sign of a flips the direction.
Q9. C. Change of base: log₇20 = ln 20/ln 7, input on top. Sanity check: 7 < 20 < 49 means the answer is between 1 and 2, and ln 20/ln 7 ≈ 3.0/1.9 ≈ 1.5 ✓. A) inverts the quotient (≈ 0.65 — visibly less than 1, impossible). B) commits the forbidden move log(M)/log(N) = log(M/N) in reverse. D) turns the quotient of logs into a difference inside. E) drops the logs entirely. Fix: log_b c = (log of the input)/(log of the base) — when in doubt, size-check against integer powers of the base.
Q10. A. Condensing enlarged the domain: the product x(x − 3) is positive at x = −2, but the original equation needs each log defined, and log(−2) does not exist — so −2 is extraneous and only x = 5 survives (log 5 + log 2 = log 10 = 1 ✓). B) validates against the condensed equation, which is exactly where ghosts hide. C) rejects a value that checks perfectly. D) has sign-flipped roots of the wrong quadratic. E) quadratics guarantee at most two candidates, not two valid solutions. Fix: after solving any log equation, substitute every candidate into the ORIGINAL equation and discard any that makes an inside nonpositive.
Q11. C. Bracket 50 between powers of 3: 3³ = 27 < 50 < 81 = 3⁴, so 3 < log₃50 < 4. A) brackets with 3¹ = 3 and 3² = 9, stopping before reaching 50. B) checks 3³ = 27 < 50 but never tests the next power, assuming the log stays below 3. D) overshoots by comparing against 3⁴ = 81 and reading the inequality backwards. E) estimates by division (50/9 ≈ 5.6) rather than by exponents. Fix: list powers of the base until they straddle the input — the two exponents that trap it are the answer.
Q12. C. The unknown t is trapped in the exponent; dividing by 0.05 cannot free it. Take the natural log of both sides: 0.05t = ln 2, so t = (ln 2)/0.05 ≈ 13.9 years. A) treats e^(0.05t) as if it were 0.05t — dividing works only on linear equations. B) is a false verification: e^(0.05·40) = e² ≈ 7.39, not 2. D) applies e instead of ln, compounding the confusion. E) uses the common log where the base is e; log₁₀2 ≈ 0.301 gives a wrong answer that looks plausible. Fix: when the unknown is in an exponent with base e, apply ln to both sides — ln e^(u) = u — then divide.