Mini-Lesson

Exponentials and logarithms

Topic 6 covers the graphs of y = aˣ and y = eˣ, why e is special (its gradient equals its value), the laws of logarithms, ln x as the inverse of eˣ, solving equations of the form aˣ = b, exponential growth and decay models, and the examiner favourite: reducing non-linear data to y = mx + c using logs.

Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.

Exponentials · the number e

Exponential functions and e

y = aˣ (a > 0) always passes through (0, 1), never touches the x-axis, and grows (a > 1) or decays (0 < a < 1).

d/dx (eˣ) = eˣe ≈ 2.71828… is the unique base whose gradient function is itself

d/dx (e^(kx)) = k e^(kx) — the chain rule brings the k down.
y = e^(−x) decays: it passes through (0, 1) and tends to 0 as x → ∞ (the x-axis is an asymptote).
eˣ > 0 for every real x — an exponential is never zero or negative.

Exam trap: e^(3x) differentiates to 3e^(3x), not 3xe^(3x−1). The power rule does not apply — the variable is in the exponent.

Quick check

Differentiate the exponential

?What is d/dx of e^(3x)?

Logarithms · definition

Logarithms — the inverse operation

A logarithm answers the question “what power?”

a^x = b ⇔ x = log_a bln x means log_e x — the natural logarithm, the inverse of eˣ

log_a a = 1 · log_a 1 = 0 · a^(log_a x) = x · ln(eˣ) = x
y = ln x is the reflection of y = eˣ in the line y = x: domain x > 0, and it cuts the x-axis at (1, 0).
You cannot take the log of a negative number or of zero.

Worked example

log₂ 32: what power of 2 gives 32? 2⁵ = 32, so log₂ 32 = 5.

e^(ln 7) = 7 — the functions undo one another.

Calculate

Your turn — evaluate a log

1Find the exact value of log₂ 32.

Set-up

2⁵ = 32, so log₂ 32 = 5.

Hint: Ask: 2 to what power gives 32?

Logarithms · laws

The laws of logarithms

log a + log b = log(ab)log a − log b = log(a/b) · log(aⁿ) = n log a · log_a a = 1

Worked example — condense

2 log x − log y = log(x²) − log y = log(x²/y).

Worked example — solve log₃(2x + 1) − log₃(x − 1) = 1

log₃[(2x + 1)/(x − 1)] = 1 ⇒ (2x + 1)/(x − 1) = 3¹ = 3

2x + 1 = 3x − 3 ⇒ x = 4. Check: log₃ 9 − log₃ 3 = 2 − 1 = 1 ✓ (and x = 4 keeps both logs positive).

Exam trap: log(a + b) is not log a + log b. And always check your solutions — a root that makes any log argument zero or negative must be rejected.

Quick check

Log laws

?Which statement is always true (for positive a, b)?

Logarithms · equations

Solving a^x = b

Take logs of both sides — any base, but ln is standard — then use the power law to bring the x down.

3^x = 20 ⇒ x ln 3 = ln 20 ⇒ x = ln 20 / ln 3= 2.9957… / 1.0986… = 2.7268…

Worked example — solve 3^x = 20 to 3 significant figures

ln both sides: x ln 3 = ln 20

x = ln 20 ÷ ln 3 = 2.995732 ÷ 1.098612 = 2.72683… = 2.73 (3 s.f.)

Check: 3^2.7268 = 20.00 ✓

Exam trap: ln20/ln3 is not ln(20/3). Divide the two logarithms — do not subtract them.

Calculate

Your turn — solve an exponential equation

2Solve 3^x = 20, giving your answer to 3 significant figures.

Set-up

x = ln 20 / ln 3 = 2.995732 / 1.098612 = 2.7268…

x = 2.73 (3 s.f.)

x =

Hint: Take ln of both sides: x ln 3 = ln 20.

Exponentials · modelling

Exponential growth and decay models

Real models look like N = N₀ e^(kt) (growth if k > 0, decay if k < 0) or P = A b^t.

Worked example — radioactive decay N = 500 e^(−0.2t), t in hours

(a) Initial amount: t = 0 gives N = 500 e⁰ = 500.

(b) After 5 hours: N = 500 e^(−1) = 500 × 0.367879 = 183.9 ≈ 184.

(c) When does it fall to 100? 500 e^(−0.2t) = 100 ⇒ e^(−0.2t) = 0.2

−0.2t = ln 0.2 = −1.60944 ⇒ t = 8.0472 ≈ 8.05 hours.

(d) Rate of decay: dN/dt = 500 × (−0.2) e^(−0.2t) = −0.2N — the rate is proportional to the amount left.

Interpretation marks: examiners ask what the constants mean. Here 500 is the initial amount and 0.2 (per hour) is the decay constant. Also expect “state a limitation of the model” — e.g. it predicts the substance never quite reaches zero.

Calculate

Your turn — evaluate the model

3A sample decays according to N = 500 e^(−0.2t), where t is in hours. Find N when t = 5, to the nearest whole number.

Set-up

N = 500 × e^(−0.2 × 5) = 500 × e^(−1) = 500 × 0.367879 = 183.94

N ≈ 184.

Hint: 500 e^(−1), and e^(−1) = 0.3679.

Calculate

Your turn — solve the model

4For the same model N = 500 e^(−0.2t), find the time t (hours, 3 s.f.) at which N = 100.

Set-up

500 e^(−0.2t) = 100 ⇒ e^(−0.2t) = 0.2

−0.2t = ln 0.2 = −1.60944 ⇒ t = 8.0472 = 8.05 hours (3 s.f.)

hours

Hint: e^(−0.2t) = 0.2, so −0.2t = ln 0.2.

Quick check

Read the graph

?Which statement about the curve y = e^(−x) is correct?

Logarithms · linearising

Reducing non-linear data to y = mx + c

The examiner gives you data, you take logs, and a straight line appears. Which log you plot depends on the model.

POWER LAW y = a xⁿ ⇒ log y = n log x + log aplot log y against log x: gradient = n, intercept = log a

EXPONENTIAL y = a bˣ ⇒ log y = (log b) x + log aplot log y against x: gradient = log b, intercept = log a

Worked example — power law

Plotting log₁₀ y against log₁₀ x gives a straight line of gradient 3 and intercept 0.5.

So n = 3 and log₁₀ a = 0.5 ⇒ a = 10^0.5 = 3.16 (3 s.f.). Model: y = 3.16 x³.

Worked example — exponential

Plotting log₁₀ y against x gives gradient 0.3010 and intercept 1.

log₁₀ b = 0.3010 ⇒ b = 10^0.3010 = 2.00; log₁₀ a = 1 ⇒ a = 10. Model: y = 10 × 2ˣ.

Exam trap: the giveaway is the axes. log y vs log x ⇒ power law. log y vs x ⇒ exponential. Get this backwards and every subsequent mark is lost.

Calculate

Your turn — power law

5Data follow y = a xⁿ. A graph of log₁₀ y against log₁₀ x is a straight line with gradient 3 and vertical intercept 0.5. Find a to 3 significant figures.

Set-up

log y = n log x + log a. So log₁₀ a = 0.5.

a = 10^0.5 = √10 = 3.1623 = 3.16 (3 s.f.), and n = 3.

a =

Hint: Intercept = log₁₀ a, so a = 10^0.5.

Calculate

Your turn — exponential model

6Data follow y = a bˣ. A graph of log₁₀ y against x is a straight line with gradient 0.3010. Find b.

Set-up

log y = (log b)x + log a, so log₁₀ b = 0.3010.

b = 10^0.3010 = 2.00 (since log₁₀ 2 = 0.30103).

b =

Hint: Gradient = log₁₀ b, so b = 10^0.3010.

Sort it

Which graph makes it linear?

Tap a model, then the graph that turns it into a straight line.

📈 log y vs log x

📉 log y vs x

➖ Already linear

Match it

Log and exponential flashcards

Tap a card on the left, then its value or equivalent.

Expression

Equals

Exponentials · exam technique

Getting full marks

Always take ln when the unknown is in the exponent; use the power law immediately.
Interpret the constants. In N = N₀e^(kt), N₀ is the initial value and k is the growth/decay rate per unit time.
Reject impossible roots. If solving gives e^x = −2, there is no solution — eˣ > 0 always.
Log-linear questions: state the model, take logs, compare with y = mx + c, then convert the gradient and intercept back.
Criticise the model. Exponential growth models eventually predict absurd values; decay models never reach exactly zero. One sentence, one mark.

Link to Topic 8: the differential equation dN/dt = kN separates to give exactly N = N₀e^(kt). Exponential models are not arbitrary — they are the solution of “the rate of change is proportional to the amount present”.

Quick check

Spot the model

?A scientist plots log₁₀ y against x and gets a straight line. Which model does this support?

Recap

The big ideas to take away

e and ln: d/dx eˣ = eˣ · ln is the inverse of eˣ · eˣ > 0 always

Log laws: log a + log b = log ab · log a − log b = log(a/b) · log aⁿ = n log a

Solving aˣ = b: take ln of both sides: x = ln b / ln a

Models: N = N₀e^(kt): N₀ is the initial value, k the rate; interpret and criticise

Linearising: y = axⁿ → log y vs log x · y = abˣ → log y vs x

Topic 6 is where algebra meets modelling — and the log-linear graph is examined almost every year. Press Finish to see your score.

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