Exponential Growth and Decay (AI SL)
Populations, radioactive samples, cooling liquids and compound interest all share the same shape: a quantity that changes by a constant proportion in each equal time step, rather than by a constant amount. This page covers how to build, read and interpret that model. It's part of the broader Exponential Models topic.
48 questions on this sub-topic.
The growth/decay model
Covered under IB syllabus reference SL2.5: exponential growth and decay models \(f(x)=ka^x+c\), \(f(x)=ka^{-x}+c\) (for \(a>0\)) and \(f(x)=ke^{rx}+c\), together with the equation of a horizontal asymptote. Typical contexts include population growth, radioactive decay, cooling of a liquid, spread of a virus, compound interest, depreciation and amortization.
Base form
\(f(x)=ka^x+c\)
\(k\) is the initial value (above the asymptote), \(a\) is the growth or decay factor, and \(c\) shifts the horizontal asymptote to \(y=c\). Not in the formula booklet - you build it from the context.
Rate form
\(f(x)=ke^{rx}+c\)
The same model written with a continuous rate \(r\): \(r>0\) gives growth, \(r<0\) gives decay. Also not in the formula booklet.
Need the full syllabus wording and formula-booklet reference table? See Exponential Models.
Worked examples
A value is modelled by \(A(t)=1500(1.04)^t.\)
(a)(i) State the initial value.
(a)(ii) State the growth factor.
(a)(iii) State the annual percentage growth.
Worked solution
of \(A(t)=1500(1.04)^t\). The coefficient is the value at \(t=0\): initial value \($1500\) A1 . The base is the growth factor: \(1.04\). A1 Since \(1.04=1+0.04\), the annual growth rate is \(4\%\) per year. A1
A sample decays as \(m(t)=80(0.5)^{t/4}\) g, \(t\) in years.
(a) State the half-life.
(b) Find the mass after 10 years, to 2 decimal places.
Worked solution
(a) Half-life. The model \(m(t)=80(0.5)^{t/4}\) halves whenever \(t\) increases by \(4\), so the half-life is \(4\) years. A1
(b) Mass after 10 years. \(10/4=2.5\): \(m(10)=80(0.5)^{2.5}=80(0.176777)\approx14.14\text{ g}.\) A1
A substance has \(M(t)=80(0.5)^{t/6}\) grams, \(t\) in hours.
(a) State the half-life.
(b) Find the mass after 18 hours.
Worked solution
(a) Half-life. The model \(M(t)=80(0.5)^{t/6}\) multiplies by \(0.5\) each time \(t\) rises by \(6\), so the half-life is \(6\) hours. A1
(b) Mass after 18 hours. \(18/6=3\) half-lives, so halve three times: \(80\to40\to20\to10\text{ g}.\) M1
Check: \(M(18)=80(0.5)^3=80(0.125)=10\) g. A1
Common mistakes
- Reading the growth/decay rate straight off the base without subtracting 1. A base of \(1.15\) is 15% growth, not 115% - always write \(a=1+r\) first before quoting a percentage.
- Sign errors when isolating the exponential term before taking logs. Rearranging \(k+c-\text{value}\) incorrectly is a common slip that produces a negative or undefined logarithm.
- Ignoring the vertical shift \(c\) when reading the asymptote. The horizontal asymptote of \(f(x)=ka^x+c\) is \(y=c\), not \(y=0\) - forgetting the shift also throws off any "initial value above the asymptote" reasoning.
Ready to practise properly?
48 growth-and-decay questions, marked instantly like the real exam.
Quick answers
How do you read the growth or decay rate off an exponential model?
Write the base as \(a=1+r\). If \(a>1\), \(r\) is the percentage growth rate; if \(a<1\) (e.g. \(a=0.95\)), the rate is \(1-a\), giving percentage decay. A base of \(1.04\) is \(4\%\) growth, not \(104\%.\)
What is half-life and how do you use it in a decay model?
Half-life is the time it takes a decaying quantity to fall to half its value. In a model like \(m(t)=80(0.5)^{t/4}\), the half-life is read directly from the exponent's divisor - here every 4 years the mass halves. See the parent topic's GDC guidance for solving these on a calculator.