Mini-Lesson
Exponentials and logarithms
This mini-lesson covers the exponential function eˣ and its inverse ln x , the laws of logarithms , solving equations of the form aˣ = b, exponential growth and decay models, and the exam favourite: using logs to turn a curve into a straight line .
Where this sits: Unit AS 1: Pure Mathematics — exponentials and logarithms: the function eˣ, ln x as its inverse, laws of logs, and exponential growth and decay models fitted using logarithms.
Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.
Definitions
Exponentials and their inverse
y = aˣ ⇔ x = log_a ya log is just an INDEX: log_a y answers "what power of a gives y?"
The special base is e = 2.71828… , chosen because the gradient of y = eˣ equals eˣ itself. Its inverse is the natural logarithm ln x = log_e x .
y = eˣ passes through (0, 1), is always positive and increases without limit.
y = ln x passes through (1, 0), is only defined for x > 0 , and is the reflection of eˣ in y = x.
e^(ln x) = x and ln(eˣ) = x — they undo each other.
ln 0 and ln(negative) do not exist. If a solution gives ln of a non-positive number, reject it.
Calculate
Read a log
1 Evaluate log₂ 32.
Check ✓
Hint: log₂ 32 asks: what power of 2 gives 32? 2⁵ = 32.
Laws
The laws of logarithms
log a + log b = log(ab) · log a − log b = log(a/b) · n log a = log(aⁿ)log_a a = 1 · log_a 1 = 0 · same base throughout!
Worked example
log₁₀ 2 + log₁₀ 50 = log₁₀(2 × 50) = log₁₀ 100 = 2
2 log 3 − log 4.5 = log 9 − log 4.5 = log(9/4.5) = log 2 = 0.301
Illegal moves: log(a + b) is not log a + log b; (log a)/(log b) is not log(a/b). Both cost marks every year.
Calculate
Log laws
2 Evaluate log₁₀ 2 + log₁₀ 50 without a calculator.
Check ✓
Hint: log a + log b = log(ab), so this is log₁₀(2 × 50) = log₁₀ 100, and 10² = 100.
Solving
Solving aˣ = b
Take logs of both sides and use the power law to bring the index down.
Worked example — 3ˣ = 20
ln(3ˣ) = ln 20 ⇒ x ln 3 = ln 20
x = ln 20 / ln 3 = 2.9957 / 1.0986 = 2.727 (3 d.p.)
Check: 3^2.727 = 20.0 ✓
Disguised quadratics: 2^(2x) − 5(2ˣ) + 4 = 0 becomes y² − 5y + 4 = 0 with y = 2ˣ, giving y = 1 or 4, so x = 0 or 2.
Calculate
Solve an exponential equation
3 Solve 3ˣ = 20. Give x to 3 decimal places.
Check ✓
Hint: Take logs: x = ln 20 ÷ ln 3 = 2.99573 ÷ 1.09861 = 2.7268…
Modelling
Exponential growth and decay
A = A₀e^(kt)k > 0 → growth · k < 0 → decay · A₀ is the initial amount (t = 0)
Worked example — half-life
A = 500e^(−0.03t). When does the amount halve, i.e. A = 250?
250 = 500e^(−0.03t) ⇒ 0.5 = e^(−0.03t)
ln 0.5 = −0.03t ⇒ t = −0.6931 / −0.03 = 23.1 (3 s.f.)
Notice the half-life ln2/0.03 does not depend on the starting amount.
Interpretation marks: be ready to say what A₀ and k mean in context , and to criticise the model (e.g. unlimited growth is unrealistic).
Calculate
Half-life
4 A radioactive sample decays according to A = 500e^(−0.03t), where t is in days. Find the time for the sample to halve, to 3 significant figures.
days
Check ✓
Hint: 0.5 = e^(−0.03t) ⇒ ln 0.5 = −0.03t ⇒ t = ln 2 ÷ 0.03 = 0.6931 ÷ 0.03.
Exam skill
Using logs to get a straight line
Two model types, two log tricks — learn which is which :
y = kxⁿ ⇒ log y = n log x + log k. Plot log y against log x : gradient n, intercept log k.y = kbˣ ⇒ log y = (log b )x + log k. Plot log y against x : gradient log b, intercept log k.
Worked example
Plotting log₁₀ y against log₁₀ x gives a straight line of gradient 1.5 and intercept 0.6.
So the model is y = kx^n with n = 1.5 and log₁₀ k = 0.6 ⇒ k = 10^0.6 = 3.98 (3 s.f.)
Calculate
Recover the constant
5 Plotting log₁₀ y against log₁₀ x gives a straight line with gradient 1.5 and vertical intercept 0.6, so y = kxⁿ. Find k to 3 significant figures.
Check ✓
Hint: The intercept is log₁₀ k, so k = 10^0.6 = 3.981…
Quick check
Which plot?
? A scientist believes y = kbˣ. Which graph should be a straight line?
y against x ❌
log y against log x ❌
log y against x ✅
y against log x ❌
Sort it
What is its value?
Tap a logarithm, then tap its value.
Quick check
Illegal move
? Which line of working is WRONG?
log 6 = log 2 + log 3 ❌
log(x + 2) = log x + log 2 ✅
3 log x = log x³ ❌
log 8 − log 2 = log 4 ❌
Match it
Match the log law
Tap the expression on the left, then its equal on the right.
Quick check
Reading the model
? A population is modelled by P = 2000e^(0.04t) with t in years. What does the 0.04 tell you?
The population starts at 40 ❌
The population grows at a continuous rate of 4% per year ✅
The population doubles every 4 years ❌
The population falls by 4% each year ❌
Quick check
Disguised quadratic
? Solve 2^(2x) − 5(2ˣ) + 4 = 0. Which are the solutions?
x = 1 or x = 4 ❌
x = 0 or x = 2 ✅
x = 2 or x = 4 ❌
no real solutions ❌
Examiner traps
Log and exponential pitfalls
log(a + b) ≠ log a + log b. There is no law for the log of a sum.(log a)/(log b) ≠ log(a/b). The quotient law needs a subtraction, not a division.Rejecting invalid solutions: ln of zero or a negative number does not exist, so discard those roots.Which plot? y = kxⁿ ⇒ log y against log x . y = kbˣ ⇒ log y against x . Mixing these up loses every mark in the part.Interpret k and the gradient in context — the model question always asks.
Recap
The big ideas to know
Definitions: y = aˣ ⇔ x = log_a y; e ≈ 2.718; ln is log to base e
Laws: log a + log b = log ab · log a − log b = log(a/b) · n log a = log aⁿ
Solving aˣ = b: take logs of both sides, then x = ln b / ln a
Models: A = A₀e^(kt) — k > 0 growth, k < 0 decay; half-life = ln2/|k|
Linearising: y = kxⁿ → plot log y vs log x · y = kbˣ → plot log y vs x
Exponentials return in differentiation (d/dx eˣ = eˣ) and in differential equation models. Press Finish to see your score.
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