Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 3: Pure Mathematics 3 · official syllabus
LOG & EXPONENTIAL LAB
Learn it by playing it
Four short missions on straightening y = ab^x data with a log plot, solving 5^x = 20 with logarithms, spotting the quadratic hiding inside e^2x - 5e^x + 6 = 0, and tuning a decay constant to a target half-life. When a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Straighten the curve
Five data points follow y = a*b^x for some constants a and b - but which? Raw y against x is a curve, hopeless to read. Plotting ln y against x turns it into a straight line. Slide the gradient and intercept until your line threads all five points, then read a and b off it.
Before you slide: to test whether data fits y = a*b^x, you plot...
Your line has gradient 1.50 and intercept 0.20. The five points sit on ln y = 0.693 + 1.099 x - nudge towards those.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Laws of logarithms and the graphs of e^x and ln x 3.2
The three laws: ln(ab) = ln a + ln b, ln(a/b) = ln a - ln b, ln(a^k) = k ln a. They are the index laws rewritten - every log rule is an exponent rule in translation.
y = e^x passes through (0, 1) with gradient 1 there, grows faster than any power of x, and has the x-axis as an asymptote as x tends to -infinity. y = ln x is its reflection in y = x: domain x > 0, through (1, 0), an asymptote on the y-axis.
e^x and ln x undo each other: e^(ln x) = x for x > 0 and ln(e^x) = x for all x. This inverse pair is what makes both solving directions possible.
Change of base: log_b(x) = ln x / ln b. Any base converts to natural logs, so ln (or log10) alone solves every exam equation.
Solving exponential equations with logarithms 3.2
For a^x = b, take logs of both sides: x ln a = ln b, so x = ln b / ln a. Worked example: 5^x = 20 gives x = ln 20 / ln 5 = 1.861, correct to 3 decimal places.
With e already in the equation, ln is the natural choice: e^(2x - 1) = 7 gives 2x - 1 = ln 7 directly.
Equations like 3^(2x) = 5^(x+1) yield to the same move: 2x ln 3 = (x + 1) ln 5, then collect x. Worked example: x(2 ln 3 - ln 5) = ln 5, so x = ln 5 / (2 ln 3 - ln 5) = 2.738, correct to 3 decimal places.
Present exact answers as log fractions first (x = ln 20 / ln 5), then round - the mark scheme carries a mark for the exact form.
Equations reducible to quadratics in e^x or a^x 3.2
Spot the pattern e^(2x) = (e^x)^2. Substitute u = e^x (or u = a^x) and the equation becomes quadratic in u.
Worked example: e^(2x) - 5e^x + 6 = 0 becomes u^2 - 5u + 6 = (u - 2)(u - 3) = 0, so u = 2 or u = 3, and x = ln 2 = 0.693 or x = ln 3 = 1.099.
Undo the substitution for every root: the u-values are not the answer. And because e^x > 0 always, a negative or zero u-root gives no solution - write that down rather than losing the mark silently.
The same pattern hides in 4^x = 2^(2x), in 9^x - 4*3^x + 3 = 0 (u = 3^x), and in log equations like (ln x)^2 - 3 ln x + 2 = 0 (u = ln x).
Reducing y = ax^n and y = ab^x to linear form 3.2
For y = a*b^x: ln y = ln a + (ln b) x. Plot ln y against x; a straight line confirms the model, the gradient is ln b and the intercept is ln a. Recover the constants with a = e^(intercept) and b = e^(gradient).
For y = a*x^n: ln y = ln a + n ln x. Plot ln y against ln x; the gradient is n itself and the intercept is ln a. Keep the two transformations apart - the exam tests choosing the right one.
Worked example: data with ln y against x giving gradient 1.099 and intercept 0.693 (ln 3 and ln 2 rounded to 3 d.p.) has b = e^(ln 3) = 3 and a = e^(ln 2) = 2 exactly, so y = 2*3^x.
Read gradient and intercept off the drawn line using two points far apart on the line, not raw data points - the line of best fit is what carries the model.
Exponential growth and decay models 3.2
N = N0 e^(kt) models growth when k > 0 and decay when written N = N0 e^(-kt) with k > 0. N0 is the starting amount - the value at t = 0.
Half-life T links to k by k = ln 2 / T, because e^(kT) must equal 2. Worked example: a sample halving in 5 years has k = ln 2 / 5 = 0.1386 (rounded to 4 d.p.) per year, so N = 800 e^(-(ln 2 / 5) t) falls to exactly 400 at t = 5.
Doubling time uses the same link: doubling in T years means k = ln 2 / T in N = N0 e^(kt).
Given two data points, solve for N0 and k by dividing the equations: the N0 cancels and one log step isolates k. State units for k (per year, per hour) - it carries the rate's meaning.
EduTribe study notes - original material aligned to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026-2027). Automated checks completed; not yet reviewed by an educator.
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