Exponential and Log Models (AI HL)
Many real quantities don't grow or shrink at a constant rate - they grow proportionally to their current size, or they climb quickly then flatten off. This page covers reading and using the two model types IB sets for these situations: exponential functions for growth and decay, and natural log functions for diminishing-returns behaviour. It's part of the broader Exponential & Logarithmic Models topic.
11 questions on this sub-topic.
The two model forms
Covered under IB syllabus reference AHL2.9: exponential models to calculate half-life, and natural logarithmic models \(f(x)=a+b\ln x\). Both are formula-booklet forms.
Exponential growth/decay model
\(f(x)=ka^x+c\) or \(f(x)=ke^{rx}+c\)
Use when a quantity changes by a constant percentage each period, approaching a horizontal asymptote \(y=c\) as \(x\to\infty\) (or \(-\infty\) for growth).
Natural logarithmic model
\(f(x)=a+b\ln x\)
Use when a quantity rises steeply at first and then levels off without a fixed ceiling - typical of learning curves and Richter-scale style comparisons. Only defined for \(x>0\).
Need the surrounding syllabus context and formula-booklet reference table? See Exponential & Logarithmic Models, including its GDC guidance.
Worked examples
The improvement in a typist's score after \(h\) hours of practice (\(h \ge 1\)) is modelled by \(S(h) = 50 + 12\ln(h)\).
(a) State the score before any additional practice (\(h=1\)).
(b) Find the score after 10 hours of practice, correct to 3 significant figures.
Worked solution
(a) \(S(1) = 50 + 12\ln(1) = 50 + 0 = 50.\) A1
(b) \(S(10) = 50 + 12\ln(10)\) M1
\(= 50 + 12(2.302585\ldots)\) A1
\(\approx 77.6.\) A1
On the Richter scale, magnitude differs by \(\log_{10}\) of amplitude ratio. How many times stronger (amplitude) is a magnitude 6.5 quake than a magnitude 5.0 one?
Worked solution
Difference \(= 6.5 - 5.0\) M1
\(= 1.5.\) A1
Amplitude ratio \(= 10^{1.5}\) M1
\(\approx 31.6\) times. A1
Common mistakes
- Forgetting the domain of a log model. A natural log model \(f(x)=a+b\ln x\) is undefined for \(x\le0\) - always state \(x>0\) when asked for the domain.
- Treating the asymptote \(c\) as zero. In \(f(x)=ka^x+c\), the constant \(c\) shifts the whole curve vertically - a decay model levels off at \(y=c\), not at \(y=0\), unless the question says otherwise.
- Mixing up which variable is inside the log. In a Richter-style comparison, it's the ratio of magnitudes that's the exponent of 10, and the ratio of amplitudes that's the log - swapping them gives an answer the wrong way round.
Ready to practise properly?
11 exponential/log-model questions, marked instantly like the real exam.
Quick answers
What is a natural logarithmic model?
A model of the form \(f(x) = a + b\ln x\), used when a quantity grows quickly at first and then levels off. It is only defined for \(x > 0\).
How is an exponential model different from a log model?
An exponential model \(f(x) = ka^x + c\) has the input in the exponent and is used for growth or decay towards an asymptote \(c\). A log model \(f(x) = a + b\ln x\) has the input inside the logarithm and models diminishing returns rather than an asymptote.