How long until a $1,000 investment at 5% per year doubles?
Solve 1000(1.05)ᵗ = 2000, i.e. (1.05)ᵗ = 2. The unknown is in the exponent — algebra's usual moves can't reach it. But you now own the tool built for exactly this: the logarithm, the exponential's inverse.
t = log₁.₀₅ 2 = ln 2 / ln 1.05 ≈ 14.2 years
Every equation in this lesson is some version of "the unknown is trapped in an exponent or inside a log — release it with the inverse."
Route 1 — matching bases (when possible): 3ˣ = 81 → 3ˣ = 3⁴ → x = 4. Fastest when both sides are powers of a common base (Lesson 11's skill).
Route 2 — take a log of both sides (always works):
4ˣ = 9 → x = log₄9 = ln 9 / ln 4 (exact) ≈ 1.585 (calculator)
Isolate the exponential first. For 5·2ˣ = 40: divide by 5 (2ˣ = 8), then apply the log. Taking log(5·2ˣ) directly works but invites errors; cleaner to strip the coefficient first.
"Exact answer" on the AP means log form (ln 9/ln 4), not a decimal.
Route 1 — rewrite as an exponential: log₂(x − 3) = 4 → x − 3 = 2⁴ → x = 19.
Route 2 — condense, then exponentiate: for multi-log equations, use the properties to get a single log first.
⚠️ The extraneous-solution check (points live here):
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 in the original equation: x = 5: log 5 + log 2 = log 10 = 1 ✓. x = −2: log(−2) is undefined ✗. Solution: x = 5 only.
Condensing can enlarge the domain (x(x−3) > 0 allows negatives that the separate logs forbid), so solving the condensed equation may produce ghosts. Always check candidates against the original.
Exponentials and logs (base > 1) are increasing, so they preserve inequality direction:
2ˣ > 16 → x > 4 (apply log₂, direction kept)ln x ≤ 2 → x ≤ e² — and the domain adds x > 0: answer 0 < x ≤ e². Forgetting the domain half is the standard error.(Base between 0 and 1 → decreasing → direction flips. Rare on the exam, but stated bases deserve a glance.)
Log functions model quantities that grow by equal amounts when the input is multiplied by equal factors — the mirror image of exponentials. Signature contexts: perceived loudness (decibels), earthquake magnitude (Richter), pH, sensation vs. stimulus. In each, multiplying the input by 10 adds a fixed amount to the output. General form f(x) = a + b·ln x, fit by 📱 LnReg (logarithmic regression) when data show rapid-then-slowing growth without a ceiling.
Plot x normally but put log(y) on the vertical axis. Magic follows:
y = a·bˣ → log y = log a + (log b)·x
Exponential data become a straight line on semi-log axes: - slope = log b (encodes the growth factor: b = 10^slope) - y-intercept = log a (encodes the initial value: a = 10^intercept)
This is both a test for exponential behavior (linear on semi-log ⇔ exponential in raw form) and a fitting technique (fit a line to (x, log y), translate back). The AP asks all three directions: raw → semi-log, semi-log → model, and "the semi-log plot is linear; what does that imply?"
[GRAPH: Two panels of the same data (x = 0…6, y = 5·2ˣ). Left: raw axes — points sweep upward in the exponential J-curve. Right: semi-log axes (vertical axis log₁₀y) — the same points fall on a straight line of slope log₁₀2 ≈ 0.301 with vertical intercept log₁₀5 ≈ 0.699. Caption: "same data, straightened."]
Problem: Solve exactly: (a) 5·2ˣ = 40 (b) 7ˣ = 30
Solution: (a) 2ˣ = 8 = 2³ → x = 3 (matching bases). (b) No common base: x = log₇30 = ln 30/ln 7 (exact form). Sanity bracket: 7¹ = 7 < 30 < 49 = 7² → x is between 1 and 2 ✓.
Interpretation: Try matching bases for ten seconds; if it doesn't crack, log both sides and move on.
Problem: Solve ln(x + 4) + ln(x − 2) = ln 27.
Solution: Condense: ln((x+4)(x−2)) = ln 27 → x² + 2x − 8 = 27 → x² + 2x − 35 = 0 → (x + 7)(x − 5) = 0 → x = −7 or x = 5. Check originals: x = 5: ln 9 + ln 3 = ln 27 ✓. x = −7: ln(−3) undefined ✗. x = 5 only.
Interpretation: The quadratic hands you two candidates; the original equation's domain (x > 2 here, from the stricter log) executes one of them. Write the check — it's a scored step.
Problem: A population grows as P(t) = 3200(1.05)ᵗ. When does it reach 6400? When does it reach 10,000?
Solution: Reach 6400 (doubling): (1.05)ᵗ = 2 → t = ln 2/ln 1.05 ≈ 14.207 years. Reach 10,000: (1.05)ᵗ = 10000/3200 = 3.125 → t = ln 3.125/ln 1.05 ≈ 1.1394/0.04879 ≈ 23.353 years.
Interpretation: Same skeleton every time: isolate the power, log both sides, divide. Notice the doubling time never depends on the 3200 — a 5% grower doubles in ≈14.2 years from any starting point.
Problem: Data plotted with log₁₀y on the vertical axis fall on the line log₁₀y = 0.3x + 1. Find the model in the form y = a·bˣ and interpret a and b. (Use 10^0.3 ≈ 2.)
Solution: Exponentiate: y = 10^(0.3x + 1) = 10¹·(10^0.3)ˣ ≈ 10·2ˣ. Initial value a = 10 (from the intercept 1: 10¹); growth factor b ≈ 2 per unit x (from the slope 0.3: 10^0.3).
Interpretation: Semi-log slope and intercept are the model's DNA in log form. Exponentiate the whole line, split with exponent rules, done.
1. (A). 81 = 3⁴ → x = 4. (C) solves 3ˣ = 27.
2. (B). Isolate: 2ˣ = 8 → x = 3. (A) stops at 2ˣ = 8 and reports 8; (D) mis-groups the division.
3. (C). x = log₄9 = ln 9/ln 4 (change of base, input on top). (B) is inverted; (D) is the quotient-inside error.
4. (D). x − 3 = 2⁴ = 16 → x = 19. (A) uses 2³; (B) computes 4 + 3; (C) solves 2ˣ = ... wrong reading.
5. (A). Candidates 5 and −2; x = −2 makes both logs undefined → {5} (worked in section (b)). (B) skips the check.
6. (B). log₂ preserves direction (increasing base): x > 4. (D) adds equality that the strict inequality forbids.
7. (C). Exponentiate: x ≤ e²; the domain demands x > 0: 0 < x ≤ e². (A) is the domain-forgetting trap.
8. (B). t = ln 2/ln 1.05 ≈ 0.693/0.0488 ≈ 14.2 years. (A) treats 5% as "1/20th per year → 20 years to double," ignoring compounding; (D) misremembers the rule of 70 as the rule of 50. (The rule-of-70 estimate 70/5 = 14 lands close to the exact answer — a good sanity check.)
9. (A). log y = log a + (log b)x — linear in x → straight line. That's the entire point of semi-log paper.
10. (D). Intercept log₁₀y = 1 at x = 0 → y = 10¹ = 10. (A) reports the log, not the value; (B) reports the slope.
11. (C). "Multiply input → add to output" is the logarithmic signature (decibels). (A) is exponential; (B) quadratic; (D) quadratic.
12. (FRQ-style, 6 points) (i) [2 pts] 96·(1/2)^(t/5) = 12 → (1/2)^(t/5) = 1/8 = (1/2)³ → t/5 = 3 → t = 15 days. (ii) [2 pts] (1/2)^(t/5) < 1/32 = (1/2)⁵ → since (1/2)^u is decreasing, the inequality flips when comparing exponents: t/5 > 5 → t > 25 days (exact); as a decimal, t > 25.000. (iii) [2 pts] 96·(1/2)^(t/5) < 0.01 → (1/2)^(t/5) < 1/9600 → t/5 > log₂9600 = ln 9600/ln 2 ≈ 9.1695/0.6931 ≈ 13.229 → t > 5·13.229 ≈ 66.144 days.
12. (FRQ-style) 🚫 A radioactive sample decays as A(t) = 96·(1/2)^(t/5), t in days. (i) Solve exactly for the time when A(t) = 12. (ii) Solve the inequality A(t) < 3, expressing the answer exactly and then as a decimal. (iii) The lab's detector cannot measure amounts below 0.01 g. Using logarithms, find how many days until the sample becomes undetectable (3 decimal places).
(a) Solve exactly: 6·3^(2x) = 54.
(b) Solve, checking for extraneous solutions: log₂(x + 3) + log₂(x − 4) = 3.
(c) The value of an account is V(t) = 800·(1.06)ᵗ dollars after t years. (i) Write an exact expression for the time at which V(t) = 2000. (ii) Between which two consecutive whole numbers of years does this time fall? Justify without a calculator. (Use: (1.06)¹⁵ ≈ 2.397 and (1.06)¹⁶ ≈ 2.540.)
(a) [2 pts] 3^(2x) = 9 = 3² [1 pt] → 2x = 2 → x = 1 [1 pt]. Check: 6·3² = 54 ✓
(b) [2 pts] log₂((x+3)(x−4)) = 3 → (x+3)(x−4) = 8 → x² − x − 12 = 8 → x² − x − 20 = 0 → (x − 5)(x + 4) = 0 → x = 5 or x = −4 [1 pt]. Check originals: x = 5: log₂8 + log₂1 = 3 + 0 = 3 ✓; x = −4: log₂(−1) and log₂(−8) undefined ✗. x = 5 only [1 pt for the check and rejection].
(c) [2 pts] (i) [1 pt] (1.06)ᵗ = 2.5 → t = ln 2.5/ln 1.06 (equivalently log₁.₀₆2.5). (ii) [1 pt] Need (1.06)ᵗ = 2.5. Given (1.06)¹⁵ ≈ 2.397 < 2.5 and (1.06)¹⁶ ≈ 2.540 > 2.5, and (1.06)ᵗ is increasing, t lies between 15 and 16 years.
1. (A). 81 = 3⁴ → x = 4. (C) solves 3ˣ = 27.
2. (B). Isolate: 2ˣ = 8 → x = 3. (A) stops at 2ˣ = 8 and reports 8; (D) mis-groups the division.
3. (C). x = log₄9 = ln 9/ln 4 (change of base, input on top). (B) is inverted; (D) is the quotient-inside error.
4. (D). x − 3 = 2⁴ = 16 → x = 19. (A) uses 2³; (B) computes 4 + 3; (C) solves 2ˣ = ... wrong reading.
5. (A). Candidates 5 and −2; x = −2 makes both logs undefined → {5} (worked in section (b)). (B) skips the check.
6. (B). log₂ preserves direction (increasing base): x > 4. (D) adds equality that the strict inequality forbids.
7. (C). Exponentiate: x ≤ e²; the domain demands x > 0: 0 < x ≤ e². (A) is the domain-forgetting trap.
8. (B). t = ln 2/ln 1.05 ≈ 0.693/0.0488 ≈ 14.2 years. (A) treats 5% as "1/20th per year → 20 years to double," ignoring compounding; (D) misremembers the rule of 70 as the rule of 50. (The rule-of-70 estimate 70/5 = 14 lands close to the exact answer — a good sanity check.)
9. (A). log y = log a + (log b)x — linear in x → straight line. That's the entire point of semi-log paper.
10. (D). Intercept log₁₀y = 1 at x = 0 → y = 10¹ = 10. (A) reports the log, not the value; (B) reports the slope.
11. (C). "Multiply input → add to output" is the logarithmic signature (decibels). (A) is exponential; (B) quadratic; (D) quadratic.
12. (FRQ-style, 6 points) (i) [2 pts] 96·(1/2)^(t/5) = 12 → (1/2)^(t/5) = 1/8 = (1/2)³ → t/5 = 3 → t = 15 days. (ii) [2 pts] (1/2)^(t/5) < 1/32 = (1/2)⁵ → since (1/2)^u is decreasing, the inequality flips when comparing exponents: t/5 > 5 → t > 25 days (exact); as a decimal, t > 25.000. (iii) [2 pts] 96·(1/2)^(t/5) < 0.01 → (1/2)^(t/5) < 1/9600 → t/5 > log₂9600 = ln 9600/ln 2 ≈ 9.1695/0.6931 ≈ 13.229 → t > 5·13.229 ≈ 66.144 days.
🎯 Exam tip: Two rituals guarantee points in this topic: (1) after solving any log equation, plug every candidate back into the original; (2) after solving any exponential equation, bracket your answer between integer powers of the base as a sanity check. Both take under fifteen seconds; both catch the exact traps the exam plants.