HookThe doubling that shut down Britain
In March 2020, UK COVID-19 cases were doubling roughly every three to four days. That phrase — "doubling every three days" — is the whole of exponential growth in five words: a quantity multiplied by a fixed factor in each equal time step. Starting from a few hundred cases, a three-day doubling reaches a thousand-fold in a month, which is why, on 23 March 2020, the government ordered a national lockdown. The maths that turned a doubling time into a policy is \(N=N_{0}e^{kt}\); the maths that reads the growth rate back out of the case data is the logarithm, its inverse.
This chapter is that pair of inverse ideas behind every growth-and-decay story. You will meet the exponential function \(a^{x}\) and its special base \(e\), and see why the gradient rule \(\tfrac{d}{dx}e^{kx}=ke^{kx}\) makes it the natural model for anything whose rate of change is proportional to its size (6.1, 6.2). You will define the logarithm as the operation that undoes exponentiation, meet \(\ln x\) (6.3), and use the log laws to manipulate them (6.4). Then you will solve equations with the unknown stuck in the exponent (6.5), use log graphs to pull hidden constants out of real experimental data (6.6), and model growth and decay while staying honest about where the model breaks (6.7) — because nothing, not even a virus, doubles forever.
ModelThe exponential function, and why the base e is special
An exponential function \(y=a^{x}\) (with \(a>0\)) has a fixed base and a variable exponent — the opposite of a power like \(x^{2}\). Every such curve passes through \((0,1)\), stays strictly positive, and has the \(x\)-axis as a horizontal asymptote. If \(a>1\) it rises ever more steeply (growth); if \(0<a<1\) it falls towards zero (decay). Its domain is all real \(x\); its range is \(y>0\).
Among all bases, \(e\approx2.71828\) is singled out by calculus: it is the base for which the curve's gradient at every point equals its own height, \(\dfrac{d}{dx}e^{x}=e^{x}\). More usefully, \(\dfrac{d}{dx}e^{kx}=ke^{kx}\), which says the rate of change is proportional to the current value. That is the exact verbal description of unrestricted population growth, radioactive decay, continuously compounded interest, and a cooling drink — a quantity whose growth or decline depends on how much of it there already is. That single property is why \(e^{kx}\), rather than any other base, is the default model throughout science.
A bacterial culture is modelled by \(N=50e^{0.2t}\), with \(t\) in hours. Because \(\dfrac{dN}{dt}=0.2\times50e^{0.2t}=0.2N\), the growth rate at any instant is one-fifth of the current population. When the culture has reached \(N=200\), it is therefore growing at \(0.2\times200=40\) cells per hour; at \(N=1000\) it is growing at \(200\) per hour. The rate is not constant — it scales with size, which is precisely what 'the gradient of \(e^{kx}\) is \(ke^{kx}\)' means and why the exponential, not a straight line, is the right model for a growing population.
ModelLogarithms — the inverse that pulls the exponent back down
A logarithm is defined as the inverse of an exponential: \(\log_{a}x=y\) means exactly \(a^{y}=x\). In words, \(\log_{a}x\) is 'the power you raise \(a\) to in order to get \(x\)'. So \(\log_{2}8=3\) because \(2^{3}=8\). Being inverses, \(\log_{a}(a^{x})=x\) and \(a^{\log_{a}x}=x\) undo each other. The graph of \(y=\log_{a}x\) is the reflection of \(y=a^{x}\) in the line \(y=x\): it passes through \((1,0)\), climbs slowly, has the \(y\)-axis as a vertical asymptote, and is defined only for \(x>0\) — you can never take the log of zero or a negative number, because no power of a positive base produces them.
The base that pairs with \(e\) is the natural logarithm, \(\ln x=\log_{e}x\), the inverse of \(e^{x}\). Hence \(\ln(e^{x})=x\), \(e^{\ln x}=x\), \(\ln1=0\) and \(\ln e=1\). Because it undoes \(e^{kt}\) in one step, \(\ln\) is the tool that extracts a growth rate \(k\) from an exponential model.
MechanismThe laws of logarithms — turning products into sums
Because logarithms are exponents, the index laws become the log laws: the product law \(\log_{a}(xy)=\log_{a}x+\log_{a}y\), the quotient law \(\log_{a}\!\big(\tfrac{x}{y}\big)=\log_{a}x-\log_{a}y\), and the power law \(\log_{a}(x^{k})=k\log_{a}x\). Their immediate consequences — \(\log_{a}1=0\), \(\log_{a}a=1\), and \(\log_{a}\!\big(\tfrac1x\big)=-\log_{a}x\) — are worth having by heart. The power law is the one that does real work: it drags an exponent down to ground level where you can solve for it.
The standing danger is to invent laws that do not exist. There is no rule for \(\log(x+y)\); the sum law is about \(\log(xy)\), not \(\log(x+y)\). Keeping the three real laws crisp — and refusing to split a log of a sum — is most of the battle.
Solve \(\log_{2}x+\log_{2}(x-2)=3\). Combine the left side with the product law: \(\log_{2}\!\big(x(x-2)\big)=3\). Undo the log by writing it in exponential form: \(x(x-2)=2^{3}=8\), so \(x^{2}-2x-8=0\), which factorises as \((x-4)(x+2)=0\), giving \(x=4\) or \(x=-2\).
Now the crucial step candidates skip: check both against the domain. \(x=-2\) would require \(\log_{2}(-2)\), which is undefined, so it must be rejected. Only \(x=4\) survives, and it checks out: \(\log_{2}4+\log_{2}2=2+1=3\). Solving the algebra without testing the domain is the single commonest way to lose the final mark here.
MechanismSolving aˣ = b — taking logs of both sides
When the unknown sits in the exponent, no amount of ordinary algebra will free it — you must take logs of both sides and then use the power law to bring the exponent down. From \(a^{x}=b\), taking natural logs gives \(x\ln a=\ln b\), so \[x=\dfrac{\ln b}{\ln a}.\] Any base works (base 10 or base \(a\) itself), but \(\ln\) is the usual choice and is expected when \(e\) is involved. The same move handles messier exponents — for \(3^{2x-1}=10\) you get \((2x-1)\ln3=\ln10\) and solve the resulting linear equation — and 'hidden quadratics' such as \(e^{2x}-5e^{x}+6=0\), which you solve by letting \(y=e^{x}\).
Return to the COVID doubling. Model cases by \(N=N_{0}e^{kt}\) with \(t\) in days, and suppose the number doubles every 3 days. Doubling means \(2N_{0}=N_{0}e^{3k}\), so \(e^{3k}=2\). Take natural logs: \(3k=\ln2\), hence \[k=\dfrac{\ln2}{3}=\dfrac{0.6931}{3}=0.231\text{ per day}.\] Now ask how long a thousand-fold rise takes: solve \(1000=e^{kt}\), so \(t=\dfrac{\ln1000}{k}=\dfrac{6.908}{0.231}=29.9\) days — about a month. A doubling time of a few days really does compress a thousand-fold surge into weeks, which is the arithmetic that forced the March 2020 lockdown.
MechanismLog graphs — reading hidden constants out of real data
Experimental data rarely arrive with their formula attached; log graphs recover it by straightening a curve into a line. Two model shapes matter. For a power law \(y=ax^{n}\), taking logs gives \(\log y=\log a+n\log x\): plotting \(\log y\) against \(\log x\) yields a straight line of gradient \(n\) and intercept \(\log a\). For an exponential law \(y=kb^{x}\), taking logs gives \(\log y=\log k+x\log b\): plotting \(\log y\) against \(x\) (not \(\log x\)) gives a line of gradient \(\log b\) and intercept \(\log k\).
The diagnostic is simple and examinable: if the data lie straight on log-log axes the underlying law is a power; if they lie straight on log-linear axes (log of \(y\) against plain \(x\)) the law is exponential. Once the line is drawn, its gradient and intercept hand you the two constants of the model.
Kleiber's law in biology relates an animal's resting metabolic rate \(B\) to its body mass \(M\) by a power law \(B=aM^{n}\). Plotting \(\log_{10}B\) against \(\log_{10}M\) for animals from a mouse to an elephant, the points fall on a straight line of gradient \(0.75\) passing through intercept \(\log_{10}B=0.70\).
Since \(\log_{10}B=\log_{10}a+n\log_{10}M\), read off the gradient \(n=0.75\) and the intercept \(\log_{10}a=0.70\), so \(a=10^{0.70}\approx5.0\). The model is therefore \(B\approx5.0\,M^{0.75}\) — the famous three-quarter-power scaling. The log-log straight line was the evidence that the law is a power (an exponential would have straightened only on log-linear axes), and the gradient and intercept delivered both constants.
CaseGrowth, decay, and the honesty a model needs
The standard models are exponential growth \(N=N_{0}e^{kt}\) (with \(k>0\)) and exponential decay \(N=N_{0}e^{-kt}\) (with \(k>0\)), where \(N_{0}\) is the value at \(t=0\) and \(k\) sets the pace. For decay, the half-life — the time to fall to half — is \(t_{1/2}=\dfrac{\ln2}{k}\), independent of where you start. These fit radioactive dating, drug clearance, cooling and unrestricted populations.
But 6.7 is really about limits, and that is where the evaluation marks live. Pure exponential growth is unbounded, whereas every real resource is finite: the COVID curve flattened as susceptible people ran out and interventions bit, so the honest refinement is the logistic model, which grows exponentially at first then levels off at a ceiling. Newton's law of cooling does not decay to zero but to room temperature, so its correct form is \(\theta=\theta_{\text{room}}+Ae^{-kt}\). A full-mark answer names such a limitation and its refinement rather than trusting the bare exponential.
Radiocarbon dating, the technique used in 1988 to date the Turin Shroud. Carbon-14 decays with a half-life of \(5730\) years, so its decay constant is \(k=\dfrac{\ln2}{5730}=1.21\times10^{-4}\) per year. A sample is found to contain \(25\%\) of its original carbon-14. Using \(N=N_{0}e^{-kt}\) with \(\tfrac{N}{N_{0}}=0.25\): \[0.25=e^{-kt}\ \Rightarrow\ kt=\ln4\ \Rightarrow\ t=\dfrac{\ln4}{1.21\times10^{-4}}=\dfrac{1.386}{1.21\times10^{-4}}\approx11\,460\text{ years}.\] A neat self-check: \(25\%\) is exactly two halvings, \(2\times5730=11\,460\) years — the two methods agree. The limitation worth stating is that the method assumes atmospheric carbon-14 has stayed constant, which calibration curves are needed to correct.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
The reflex for any equation with the unknown in the exponent is to take logs and use the power law — write the line \(x=\dfrac{\ln b}{\ln a}\) so the method mark is secured even before you reach for the calculator. When you combine logs to solve an equation, always finish by checking each solution against the domain and rejecting any that would take the log of zero or a negative number; that rejection is a mark in its own right.
On log-graph questions, state precisely which quantity is plotted against which and interpret the line correctly: a straight log-log plot means a power law \(y=ax^{n}\) with gradient \(n\), while a straight log-linear plot means an exponential \(y=kb^{x}\) with gradient \(\log b\) — mixing these up is the classic error. For growth and decay, identify \(N_{0}\) and \(k\) explicitly, use \(t_{1/2}=\tfrac{\ln2}{k}\) and cross-check whole-number half-lives against the exponential answer, and for any 6.7 modelling mark name a limitation (unbounded growth, decay that should level at a floor) together with its refinement — logistic growth or a shifted cooling model — rather than trusting the raw exponential.