AQA A-level Mathematics 7357, section F (Exponentials and logarithms), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.
Prerequisites: index laws and scientific calculators from GCSE. Each lesson develops its own method before independent transfer.
These are original teaching examples, not official AQA material.
A sample halves every 3.5 hours, but the laboratory needs the level at 5 hours — no table gives it. How does one formula replace the missing table?
Start with a prediction. State what the model y = Ae^(kt) needs as input, and how a log plot could recover k from data.
Build the model
In y = Ae^(kt), A is the initial amount and k is the proportional rate: positive for growth, negative for decay. The half-life 半衰期 is ln 2/|k|; doubling time is ln 2/k.
The logarithm is the inverse of the exponential: y = a^x means x = log_a y, defined only for positive y. The log laws follow from the index laws: log(uv) = log u + log v, log(u/v) = log u − log v, log(uᵏ) = k log u. Solving a^x = b gives x = ln b / ln a. Plotting ln y against t turns y = Ae^(kt) into the straight line ln y = ln A + kt.
half-life: The time for a decaying quantity to fall to half its initial value; logarithm: The exponent to which a stated base must be raised to produce a given positive number.
Choose evidence that can test it
A fitted exponential is a model: check its stated time unit and range before predicting, and compare a prediction with a fresh measurement (a residual) before trusting it. A negative k means decay; a negative starting amount usually contradicts the context.
In log equations, record the original domain first: every argument of every log must be positive, so candidates produced after combining logs must be checked against it.
Work from known quantities
State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
Known: y = 80e^(−0.2t), y in grams, t in hours. Half the amount means 40 = 80e^(−0.2t), so e^(−0.2t) = 1/2. Taking ln: −0.2t = −ln 2, so t = ln 2/0.2 ≈ 3.466 hours. The same k gives the half-life for any starting amount.
Example:
Solve 3^x = 20 using the same sequence: take logs → apply the power law → divide → interpret. (Answer: x ln 3 = ln 20, so x = ln 20/ln 3 ≈ 2.727. Check: 3^2.727 ≈ 20.)
Check the conclusion and its limits
Log laws apply only to positive arguments; ln(x + y) has NO sum law (ln 2 + ln 3 = ln 6 ≠ ln 5). The exponential model needs its unit of time matched to k; an effective yearly rate of 10% means k = ln 1.1, not k = 0.1.
Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Warn:
For positive x and y, ln(x + y) = ln x + ln y. This claim is false: the product law is ln(xy) = ln x + ln y; there is no law for a sum inside a logarithm, as x = y = 1 shows (ln 2 ≈ 0.69 but ln 1 + ln 1 = 0).
Key:
Exponentials and logarithms: y = Ae^(kt) with half-life ln 2/|k|; logs invert exponentials, follow the product/quotient/power laws, and turn an exponential fit into a straight line. Check domains and units before solving.
