Lesson 4: Exponents, Logarithms and Applications

Exponential functions model change that is proportional to current size — populations, radioactive decay, compound interest, cooling. Their inverses, logarithms, are among the most examined ideas in AA (Topics 1 and 2), because they test laws, equation-solving and graph sense in a single package. Two bases dominate: 10 (engineering scales: decibels, pH, Richter) and e ≈ 2.71828 (continuous growth and calculus). The natural logarithm ln x means log_e x, and the functions e^x and ln x are mutual inverses whose graphs are reflections in y = x.

The laws — and why they are true

Every logarithm law is an exponent law in disguise. From a^m × a^n = a^(m+n) comes log(xy) = log x + log y; from a^m / a^n comes log(x/y) = log x − log y; from (a^m)^n comes log(x^r) = r log x. The change of base rule, log_a x = log_b x / log_b a, is how your GDC evaluates log_7 50 with its ln or log₁₀ keys. Non-negotiable domain facts: log x exists only for x > 0, log 1 = 0, log_a a = 1, and there is no law for log(x + y) — the single most abused non-identity in the course.

Worked Example 1 — laws in both directions. (a) Write log 48 − log 3 as a single logarithm. (b) Given log₂ 3 = p, express log₂ 24 in terms of p.

Solution. (a) log 48 − log 3 = log(48/3) = log 16. (b) log₂ 24 = log₂ (8 × 3) = log₂ 8 + log₂ 3 = 3 + p. The direction matters: exams alternate between expanding and condensing, and both directions are marked step by step — write each law as its own line.

Solving exponential and logarithmic equations

The universal strategy: isolate the exponential, then take logarithms of both sides (or use the inverse function). Equations where both sides share a base dissolve instantly: 2^(x+1) = 8 = 2³ gives x + 1 = 3. Quadratics in disguise appear constantly: 4^x − 3(2^x) + 2 = 0 is a quadratic in 2^x via 4^x = (2^x)².

Worked Example 2 — two equation types (past-paper style). Solve (a) 5^(2x) = 125^(x−1); (b) log₃(x + 2) + log₃(x − 4) = 2.

Solution. (a) Both sides are powers of 5: 125 = 5³, so 5^(2x) = 5^(3x−3), giving 2x = 3x − 3 and x = 3. (b) Condense: log₃[(x + 2)(x − 4)] = 2, so (x + 2)(x − 4) = 3² = 9, i.e. x² − 2x − 8 = 9, i.e. x² − 2x − 17 = 0, giving x = 1 ± 3√2 ≈ 5.24 or −3.24. Now the domain check that earns the final mark: log₃(x − 4) requires x > 4, so x ≈ −3.24 is rejected (it also makes log₃(x + 2) undefined since x + 2 < 0). Answer: x = 1 + 3√2 ≈ 5.24. On Paper 2 the GDC graph intersection finds the root in seconds, but the rejection reasoning must still be written.

Worked Example 3 — the exponential model and its inverse. A bacterial culture starts at 500 cells and doubles every 3 hours, N(t) = 500 × 2^(t/3). (a) Find the population after 10 hours. (b) Find when it reaches 100 000 cells.

Solution. (a) N(10) = 500 × 2^(10/3) ≈ 500 × 10.079 ≈ 5040 cells (3 s.f.). (b) Solve 500 × 2^(t/3) = 100 000 → 2^(t/3) = 200 → t/3 = log₂ 200 = ln 200 / ln 2 ≈ 7.6439, so t ≈ 22.9 hours. Method note: Paper 1 demands the logarithm step shown; Paper 2 permits the GDC solver, but you must define the equation you solved. Half-life and doubling-time questions are this example with the roles of given and unknown swapped.

Graphs: e^x, ln x and transformations

f(x) = a^x (a > 1) grows without bound, has horizontal asymptote y = 0, and passes through (0, 1) — every pure exponential does, a fact worth one mark more often than you would expect. Its inverse ln x has vertical asymptote x = 0 and passes through (1, 0). Interest applications use A = P(1 + r/k)^(kt) for k compoundings per year, and the continuous limit A = Pe^(rt); IB financial questions are routine marks *if* the exponent counts the total number of periods, not years.

Common mistakes

• Inventing laws: log(x + y) = log x + log y is false; log(x^r) = r log x works only on a single term.
• Losing solutions or keeping illegal ones in log equations — always check the domain of every original logarithm.
• Solving 2^x = 10 as x = 5 (dividing instead of logging).
• ln and e^x not used as inverses: e^(ln x) = x only for x > 0; ln(e^x) = x for all x.
• Interest exponent errors: monthly compounding for 3 years is 36 periods, not 3.
• Early rounding inside the exponent — keep full precision until the final line, then three significant figures.

IB Exam Experience

Question 1 [2 marks]. Solve 3^(x−1) = 27√3, giving an exact answer. *Marking notes:* M1 for writing 27√3 = 3^(7/2) (27 = 3³, √3 = 3^(1/2)); A1 for x − 1 = 7/2, x = 9/2. Exact form required; a decimal forfeits the accuracy mark on Paper 1.

Question 2 [3 marks]. Given ln 5 = a and ln 3 = b, express ln 45 − ln 25 in terms of a and b. *Marking notes:* M1 for ln(45/25) = ln(9/5) or expanding each term; M1 for prime structure 45 = 3² × 5 and 25 = 5²; A1 for 2b − a. (Check: ln 45 − ln 25 = (2b + a) − 2a = 2b − a. ✓)

Question 3 [4 marks]. Solve log₂ x + log₂(x − 2) = 3. *Marking notes:* M1 condensing to log₂[x(x − 2)] = 3; M1 converting to x² − 2x = 8; A1 for x = 4 or x = −2; A1 for rejecting x = −2 because both original logarithms require x > 2. The rejection mark is the one candidates throw away.

Extended problem (Paper 2 style) [10 marks]. A car bought for $24 000 depreciates by 12% per year. (a) Write the value model V(t). [2] (b) Find its value after 6 years. [2] (c) Find when its value first falls below $8 000. [3] (d) A second model uses continuous decay V(t) = 24000 e^(−kt). Find k so the models agree after one year, and comment on which predicts the lower value after 20 years. [3] *Marking notes:* (a) M1 A1: V(t) = 24000(0.88)^t. (b) A2: V(6) ≈ $11 100 (24000 × 0.88⁶ = 24000 × 0.4644 ≈ 11 145 → 11 100 to 3 s.f.). (c) M1 for 0.88^t = 1/3 or a GDC intersection; M1 for logs: t = ln(1/3)/ln(0.88) ≈ 8.59; A1 for interpretation: during the 9th year / after ≈ 8.59 years — "first falls below" rounding direction is the examiner's focus. (d) M1: e^(−k) = 0.88 → k = −ln 0.88 ≈ 0.128; A1: with equal one-year decay the models are identical in form (0.88^t = e^(t ln 0.88)); A1 for spotting that agreement at year 1 forces agreement everywhere — a reasoning mark for noticing, not computing.

Exploration (IA) idea: track the cooling of a cup of tea with a thermometer app, fit Newton's law of cooling (an exponential plus an asymptote), and evaluate the model against data — a textbook-strong IA combining modelling, logs to linearise, and honest evaluation.

Going deeper: linearising with logarithms

Plotting ln y against x (for exponential relationships y = ab^x) or ln y against ln x (for power laws y = ax^n) turns curves into straight lines whose gradient and intercept reveal the parameters — the bridge between this lesson and regression in Lesson 16, and a favourite Paper 2 modelling structure.

Extra worked example. Data suggests N = N₀e^(kt) for a cooling experiment. At t = 2, N = 84.3; at t = 7, N = 40.6, with N₀ = 100. Estimate k, then predict when N = 10.

Solution. Take ln: ln N = ln 100 + kt. From the first point: ln 84.3 = 4.6052 + 2k → k = (4.4348 − 4.6052)/2 ≈ −0.0852. The second point gives k = (ln 40.6 − 4.6052)/7 = (3.7038 − 4.6052)/7 ≈ −0.1288. The two estimates disagree — real data always does this — and a regression of ln N on t over all the data points (Lesson 16) is the way to balance them. Using the average of the two-point estimates k ≈ −0.107: N = 10 gives t = ln(0.1)/k = (−2.3026)/(−0.107) ≈ 21.5. The honest lesson: two-point fits are fragile; Lesson 16's least-squares line on the linearised data is the professional version — and saying so in a solution is exactly the modelling commentary examiners reward.

Extra exam practice [4 marks]. Solve ln(x − 1) + ln(x + 2) = ln 4, stating the domain first. *Marking notes:* M1 for the domain: x > 1 (both arguments positive); M1 for condensing: ln[(x − 1)(x + 2)] = ln 4 → x² + x − 2 = 4; A1 for x = 2 or x = −3; A1 for rejecting −3 by the domain, answer x = 2.

Going further: mixed bases and half-life fluency

Two refinements separate competent log work from exam-grade log work. First, mixed-base equations: convert everything to one base with the change-of-base rule. Solve log₂x + log₄x = 6: since log₄x = log₂x / log₂4 = (log₂x)/2, the equation becomes (3/2) log₂x = 6, so log₂x = 4 and x = 16. Check: log₂16 + log₄16 = 4 + 2 = 6 ✓. Second, half-life as a power of 2: a sample of 80 g with half-life 5 days has mass m(t) = 80 × 2^(−t/5); after 12 days, m = 80 × 2^(−2.4) ≈ 80 × 0.189 ≈ 15.2 g (3 s.f.). Writing decay as a power of 2 — rather than reaching for e and a fitted k — keeps Paper 1 solutions exact until the final evaluation, and the same template handles doubling times, drug clearance and depreciation: identify the halving (or doubling) period, exponentiate in units of that period, evaluate last.

Extra exam practice [3 marks]. Solve 3^x = 5^(x − 1). *Marking notes:* M1 for taking natural logs of both sides: x ln 3 = (x − 1) ln 5; M1 for expanding the bracket and collecting x on one side: x ln 5 − x ln 3 = ln 5, so x(ln 5 − ln 3) = ln 5; A1: x = ln 5 / (ln 5 − ln 3) ≈ 1.609/0.511 ≈ 3.15. The sign discipline in the collection step is the whole question.