Lesson 3: Exponential and Logarithmic Models

Exponential and logarithmic functions are inverses of each other, and between them they model anything that grows or shrinks by a percentage: populations, investments, radioactive samples, sound and earthquake intensity. Precalculus treats them as one topic because every technique is the same move in two directions — exponentiate to undo a logarithm, take a logarithm to undo an exponential.

Exponential growth and decay

The model is A(t) = A0(1 + r)^t for growth at rate r per period, and A(t) = A0(1 − r)^t for decay; with a half-life h it is A(t) = A0(1/2)^(t/h). Worked example: $500 invested at 4% annual interest gives A(t) = 500(1.04)^t, so A(10) = 500(1.4802) ≈ $740.12. The base 1.04 is the growth factor: each year multiplies the balance by 1.04. A table makes the multiplicative structure visible — 500.00, 520.00, 540.80, 562.43 — the *increments* grow (20, 20.80, 21.63), unlike linear growth where increments are constant. That distinction — constant ratio versus constant difference — is how you recognise exponential data in Lesson 19.

Doubling time and half-life

Setting A(t) = 2A0 in 500(1.04)^t gives (1.04)^t = 2, so t = ln 2/ln 1.04 ≈ 17.7 years: the doubling time depends only on the rate, not the principal. Worked example — carbon dating. Carbon-14 has a half-life of about 5,730 years. A 40 g sample left for 10,000 years holds A = 40(1/2)^(10000/5730) = 40(1/2)^1.745 ≈ 40(0.298) ≈ 11.9 g. Running the model backwards — measuring 11.9 g today to infer 10,000 years — is exactly how archaeologists date artefacts, and it is a logarithm problem in disguise.

Logarithms: undoing the exponential

log_b(y) = x means b^x = y. The laws mirror the exponent laws: log(uv) = log u + log v, log(u/v) = log u − log v, log(u^p) = p·log u. Worked example — solving with logs. Solve log2(x) + log2(x − 2) = 3. Combine: log2(x(x − 2)) = 3, so x(x − 2) = 2^3 = 8, giving x^2 − 2x − 8 = 0 and x = 4 or x = −2. But log2(−2) is undefined — the domain of the original equation requires x > 2 — so the only solution is x = 4. Logs of sums do not split (log(a + b) has no law), a fact the mistakes section returns to.

Solving exponential equations

When both sides share a base, equate exponents. When they do not, take logarithms. Worked example: 3^(2x − 1) = 20. Take ln of both sides: (2x − 1) ln 3 = ln 20, so 2x − 1 = ln 20/ln 3 ≈ 2.7268 and x ≈ 1.863. Check: 3^2.727 ≈ 20.02 ✓ (rounding). For base e, ln is its own undo button: e^(2x) = 7 gives x = ln 7/2 ≈ 0.973. The change-of-base formula log_b(y) = ln y/ln b is what makes all of this computable on a calculator that only has ln and log10.

Modelling with logarithms: pH and Richter

Logarithms compress enormous ranges. pH: pH = −log10[H+]. For [H+] = 3.2 × 10^−5, pH = −log10(3.2 × 10^−5) ≈ 4.49 — mildly acidic. Because the scale is logarithmic, one pH unit means a factor of 10 in acidity. Richter scale: magnitudes are log10 of amplitude, so a magnitude 7.5 quake has 10^1.5 ≈ 31.6 times the amplitude of a magnitude 6.0 — "1.5 bigger" is really "about 32 times bigger." Whenever a quantity spans many orders of magnitude, expect a logarithm in its model.

Common mistakes

• Splitting log(a + b). There is no law: log2(8 + 8) = log2 16 = 4, but log2 8 + log2 8 = 6. Only products, quotients and powers split.
• Forgetting to reject extraneous roots. Combining logs can enlarge the apparent domain; x = −2 above satisfies the combined equation but not the original. Always check candidates in the *original* equation.
• Cancelling the log like a factor. ln(x)/ln(3) is not x/3; the logarithm is a function, not a multiplier.
• Mixing (1 + r)^t with continuous compounding. A0·e^(rt) uses a rate quoted for continuous growth; the two models agree only after converting r. Read the problem's wording.
• Misplacing the exponent when solving. From 5·2^x = 40, first isolate: 2^x = 8, x = 3. Taking logs before isolating produces ln 5 + x ln 2 = ln 40 — legal but messier, and a common source of sign errors.

Exam Experience

1. If 2^(x + 1) = 24, then x ≈

• (A) 4.58
• (B) 3.58
• (C) 2.58
• (D) 5.58

Answer: (B). x + 1 = log2 24 ≈ 4.585, so x ≈ 3.585. (A) is log2 24 itself — the answer if you forget the +1 in the exponent, which is exactly the trap the question sets.

2. The solution of log3(x) + log3(x − 8) = 2 is

• (A) x = 9 only
• (B) x = −1 only
• (C) x = 9 or x = −1
• (D) no solution

Answer: (A). Combining gives x(x − 8) = 9, i.e. x^2 − 8x − 9 = 0, with roots 9 and −1. But log3(x − 8) requires x > 8, so −1 is extraneous — (C) is the mark of a student who skipped the domain check.

3. A population doubles every 12 years. Starting from 3,000, how long until it reaches 48,000?

• (A) 24 years
• (B) 36 years
• (C) 48 years
• (D) 60 years

Answer: (C). 48,000/3,000 = 16 = 2^4, so four doublings are needed: 4 × 12 = 48 years. Counting doublings beats setting up the exponential equation when the ratio is a power of 2.

Constructed response. A bacterial colony starts with 200 cells and triples every 5 hours, so N(t) = 200·3^(t/5), with t in hours. (a) How many cells are present after 12 hours? (b) At what time does the colony reach 10,000 cells? Give your answer to the nearest tenth of an hour. (c) Explain why the answer to (b) requires a logarithm even though the model is exponential.

Scoring guide. (a) 2 points: N(12) = 200·3^2.4 — 1 point for the substitution; 3^2.4 ≈ 13.97 so N ≈ 2,793 cells — 1 point. (b) 2 points: 3^(t/5) = 50, so t/5 = ln 50/ln 3 ≈ 3.561 and t ≈ 17.8 hours — 1 point for isolating the power and applying a logarithm, 1 point for the value. (c) 1 point: the unknown t sits in the exponent, and only the inverse operation — the logarithm — can bring it down to be solved for; evaluating the model (part a) needs no inverse because t is given.

Going deeper: continuous growth, decibels and a disguised quadratic

Continuous compounding replaces (1 + r)^t by e^(rt): A(t) = A0·e^(rt). Worked example: $1,000 at 6% for 10 years. Continuous: 1000·e^0.6 ≈ 1000 × 1.8221 = $1,822.12. Annual compounding: 1000 × 1.06^10 ≈ $1,790.85. Continuous compounding always wins (interest earns interest instantly), and e is not decorative here — it is the limit of (1 + r/m)^(mt) as the compounding frequency m → ∞, another limit hiding inside a formula.

Decibels use a log scale with a factor of 10: β = 10·log10(I/I0), where I0 = 10^−12 W/m² is the hearing threshold. A sound of intensity 10^−6 W/m² has β = 10·log10(10^6) = 60 dB — normal conversation. Doubling intensity adds 10·log10 2 ≈ 3 dB, whatever the starting level: equal *ratios*, equal decibel steps.

A disguised quadratic. Solve e^(2x) − 3e^x + 2 = 0. Substitute u = e^x: u^2 − 3u + 2 = (u − 1)(u − 2) = 0, so e^x = 1 or e^x = 2, giving x = 0 or x = ln 2 ≈ 0.693. Both are valid (e^x > 0 always, so no extraneous roots can appear — but if u = −1 had appeared, it would have to be rejected since e^x is never negative). Whenever an equation is a polynomial in e^x, ln x, or sin x, the method is identical: substitute, solve the polynomial, substitute back, and audit the answers against the inner function's range. This single pattern solves a large fraction of all "hard" exponential and trig equations.

Workshop: growth to a target, and a log equation

Reaching a target population. Bacteria double every 40 minutes from 300 cells: N(t) = 300·2^(t/40). After 3 hours (t = 180): N = 300·2^4.5 = 300 × 22.63 ≈ 6,788 cells. When does the culture reach 100,000? 2^(t/40) = 100000/300 ≈ 333.3, so t/40 = log2(333.3) ≈ 8.38 and t ≈ 335 minutes ≈ 5.6 hours. Note the shape of the answer: reaching 333 times the starting count takes only about 8.4 doubling periods — logarithms convert intimidating growth into a count of doublings, which is why "how many doublings?" is the professional's first question about any exponential.

A log equation with a domain trap. Solve ln(x − 1) + ln(x + 1) = ln 8. Combine: ln(x^2 − 1) = ln 8 → x^2 − 1 = 8 → x = ±3. The domain requires x − 1 > 0, i.e. x > 1, so x = 3 only; x = −3 satisfies the combined equation but neither logarithm exists there. The pattern of Lesson 3 — combine, solve, audit — is mechanical once the audit is habitual.

Final drills: bases and inequalities

Two closing skills. Change of base for exact values: log_4(32) = ln 32/ln 4 — but recognise 32 = 4^(5/2) (since 4^2 = 16 and 4^(1/2) = 2, so 4^(5/2) = 16 × 2 = 32), giving exactly 5/2. Half-integer logarithms appear whenever the argument is a root of the base; check for them before reaching for decimals. Exponential inequalities: (1/2)^x > 8 becomes 2^(−x) > 2^3, and since the base-2 exponential is increasing, −x > 3, i.e. x < −3 — note the flip comes from the negative exponent, and bases between 0 and 1 reverse monotonicity directly: (1/2)^x is decreasing, so the inequality direction reverses when comparing exponents. State the monotonicity you are using; that sentence is where the marks live.