1 exam-style question plus unlimited generated ones on this topic, at up to four difficulty levels, with full mark schemes. Notes and worksheets are open to everyone. Practice needs a free account.
Or take a quick quiz: 11 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
Taking logarithms turns the curves \(y = ax^n\) and \(y = kb^x\) into straight lines, so you can find the constants from the gradient and intercept of a line. Questions give data or a line of best fit, often in a real context, and ask you to find and interpret the constants.
Key facts
\(y = ax^n \Rightarrow \log y = n\log x + \log a\)
\(\log y\) against \(\log x\): gradient \(n\), intercept \(\log a\)
\(y = kb^x \Rightarrow \log y = x\log b + \log k\)
\(\log y\) against \(x\): gradient \(\log b\), intercept \(\log k\)
Undo \(\log_{10}\) with \(10^{(\ldots)}\); undo ln with \(\mathrm{e}^{(\ldots)}\)
Straight line for \(\log y\) against \(\log x\): power law. Straight line for \(\log y\) against \(x\): exponential
Notes
Power laws: \(y = ax^n\)
Take logs of both sides (base 10 or base e): \(\log y = \log a + n\log x\)
Compare with \(Y = mX + c\), where \(Y = \log y\) and \(X = \log x\): the gradient \(m\) is \(n\) and the intercept \(c\) is \(\log a\)
So if \(y = ax^n\), the graph of \(\log y\) against \(\log x\) is a straight line.
Exponential laws: \(y = kb^x\)
Take logs: \(\log y = \log k + x\log b\)
Here \(Y = \log y\) and \(X = x\) (not \(\log x\)): the gradient is \(\log b\) and the intercept is \(\log k\)
So if \(y = kb^x\), the graph of \(\log y\) against \(x\) is a straight line.
Finding and using the constants
diagram
Find the gradient from two points on the line (points on a line of best fit need not be data points).
Left: a power law gives a straight line of \(\log y\) against \(\log x\). Right: an exponential law gives a straight line of \(\log y\) against \(x\)
The intercept is the value of \(\log y\) where the horizontal variable is 0. Check that the horizontal axis really starts at 0.
Undo the logs: with \(\log_{10}\), \(a = 10^{c}\) and \(b = 10^{m}\); with ln, \(a = \mathrm{e}^{c}\) and \(b = \mathrm{e}^{m}\). For a power law, \(n\) is the gradient itself.
If you are given the equation of the line, e.g. \(\log_{10} V = 4.3 - 0.1t\), read off \(m\) and \(c\) directly.
In context, \(k\) is the starting value (when \(x = 0\)) and \(b\) is the multiplier for each increase of 1 in \(x\). Predictions outside the range of the data (extrapolation) are unreliable.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Exponential model from a log graph
5 to 7 marksGrade B
Take logs of the model to show it gives a straight line: \(\log V = \log k + t\log b\)
Find the gradient and the intercept of the line.
Undo the logs: \(k = 10^{\text{intercept}}\) and \(b = 10^{\text{gradient}}\)
Interpret \(k\) and \(b\) in context.
Example. The value, \(V\) pounds, of a car \(t\) years after it was bought is modelled by \(V = kb^t\), where \(k\) and \(b\) are constants. The graph of \(\log_{10} V\) against \(t\) is a straight line through the points \((0, 4.3)\) and \((5, 3.8)\). (a) Show that \(\log_{10} V = \log_{10} k + t\log_{10} b\). (b) Find the values of \(k\) and \(b\), giving each to 3 significant figures. (c) Interpret the value of \(b\) in this context.
Show the model answerHide the model answer
(a)\[\begin{aligned}[b]\log_{10} V &= \log_{10} k + \log_{10} b^t \\ &= \log_{10} k + t\log_{10} b\end{aligned}\]
B1
(b) Gradient \(= \dfrac{3.8 - 4.3}{5} = -0.1\)
M1
\(\log_{10} k = 4.3 \Rightarrow k = 10^{4.3} = 20\,000\) (3 s.f.)
M1 A1
\(\log_{10} b = -0.1 \Rightarrow b = 10^{-0.1} = 0.794\) (3 s.f.)
A1
(c) Each year the car's value is multiplied by 0.794, so it loses about 20.6% of its value each year
B1
Power law from a log–log graph
4 to 6 marksGrade B
Take logs: \(\ln y = n\ln x + \ln a\), so the gradient is \(n\) and the intercept is \(\ln a\)
Work out the gradient from two points on the line.
Find the intercept by substituting a point into \(\ln y = n\ln x + c\)
Undo the log, \(a = \mathrm{e}^{c}\), then use the model.
Example. Two quantities \(x\) and \(y\) are thought to satisfy \(y = ax^n\), where \(a\) and \(n\) are constants. A graph of \(\ln y\) against \(\ln x\) is a straight line through \((1, 3.5)\) and \((3, 2.5)\). (a) Find the value of \(n\) and the value of \(a\), giving \(a\) to 3 significant figures. (b) Use the model to estimate \(y\) when \(x = 9\)
Show the model answerHide the model answer
(a) \(\ln y = n\ln x + \ln a\); gradient\[\begin{aligned}[b]n &= \frac{2.5 - 3.5}{3 - 1} \\ &= -0.5\end{aligned}\]
Say which quantities to plot: \(\log y\) against \(\log x\) for a power law, or \(\log y\) against \(x\) for an exponential.
Say that the model fits if the points lie close to a straight line.
Say how the gradient and the intercept give the constants.
Example. A scientist thinks that two variables \(P\) and \(t\) are related by \(P = at^n\), where \(a\) and \(n\) are constants. She has a table of values of \(P\) and \(t\). Explain how she could use a graph to test her idea and to estimate \(a\) and \(n\)
Show the model answerHide the model answer
Plot \(\log_{10} P\) against \(\log_{10} t\)
B1
If the points lie close to a straight line, the model is suitable
B1
The gradient of the line gives \(n\)
B1
The intercept on the vertical axis is \(\log_{10} a\), so \(a = 10^{\text{intercept}}\)
B1
Don’t lose marks
Giving \(k = 4.3\) instead of \(k = 10^{4.3}\): forgetting to undo the log.
Undoing with the wrong base, e.g. using \(\mathrm{e}^{c}\) when the axis is \(\log_{10}\)
Reading the intercept from a vertical axis that is not drawn at 0 on the horizontal scale.
Plotting \(\log y\) against \(\log x\) for an exponential model, or \(\log y\) against \(x\) for a power law.
Rounding the gradient or intercept before raising 10 (or \(\mathrm{e}\)) to that power.
More tips
Memory tricks
Count the logs: only \(y\) logged means exponential (\(y = kb^x\)); both logged means power law (\(y = ax^n\)).
The intercept is always the log of the constant in front (\(\log a\) or \(\log k\)), so it always needs undoing.
For a power law the gradient is \(n\) itself; for an exponential the gradient is \(\log b\), so it needs undoing too.
Sense check: a line sloping down in an exponential model means \(0 \lt b \lt 1\), which is decay.
Exam technique
In 'show that' parts, show the log laws step by step: \(\log(kb^t) = \log k + \log b^t = \log k + t\log b\)
Give the constants to the accuracy asked for, usually 3 significant figures.
When asked to interpret, use the context and units, e.g. '\(k\) is the value of the car when it was new'.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade D
Take logs of \(y=ax^n\)\(\log y = n\log x + \log a\): a straight line when \(\log y\) is plotted against \(\log x\)
Grade D
Take logs of \(y=kb^x\)\(\log y = x\log b + \log k\): a straight line when \(\log y\) is plotted against \(x\)
Grade C
Choose the right axes for a straight linePower law: \(\log y\) against \(\log x\). Exponential: \(\log y\) against \(x\)
Grade B
Find \(a\) and \(n\) from a lineThe gradient is \(n\); the intercept \(c\) is \(\log a\), so \(a = 10^{c}\) (or \(a = \mathrm{e}^{c}\) if the axes use ln).
Grade B
Find \(k\) and \(b\) from a lineThe intercept \(c\) is \(\log k\) and the gradient \(m\) is \(\log b\), so \(k = 10^{c}\) and \(b = 10^{m}\)
Grade A
Interpret the constants and criticise the modelE.g. \(k\) is the value when \(x = 0\) and \(b\) is the multiplier per unit of \(x\); predictions far outside the data are unreliable.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy7 marks
The variables \(x\) and \(y\) are related by \(y = ax^{n}\), where \(a\) and \(n\) are constants.
(a) Show that \(\log_{10} y = \log_{10} a + n\log_{10} x\).[2]
(b) A graph of \(\log_{10} y\) against \(\log_{10} x\) is a straight line with gradient \(2.5\) that meets the vertical axis at \((0, 0.45)\). Find the value of \(n\) and the value of \(a\), giving \(a\) to 3 significant figures.[3]
(c) Use the model to find the value of \(y\) when \(x = 8\), giving your answer to 3 significant figures.[2]
Show the answer and mark scheme
(a)Answer: Shown
M1 for taking logarithms of both sides and using the addition law: \(\log_{10} y = \log_{10} a + \log_{10} x^{n}\)
A1* for using the power law to reach the given result
Worked solution: \(\log_{10} y = \log_{10}(ax^{n}) = \log_{10} a + \log_{10} x^{n} = \log_{10} a + n\log_{10} x\)
(b)Answer: \(n = 2.5, \ a = 2.82\)
B1 for \(n = 2.5\)
M1 for \(\log_{10} a = 0.45\)
A1 for \(a = 2.82\)
Worked solution: Comparing with \(Y = mX + c\): the gradient gives \(n = 2.5\) and the intercept gives \(\log_{10} a = 0.45 \Rightarrow a = 10^{0.45} = 2.82\).
(c)Answer: \(y = 510\)
M1 for \(2.82 \times 8^{2.5}\) (or \(\log_{10} y = 0.45 + 2.5\log_{10} 8\))
A1 for awrt \(510\)
Worked solution: \(y = 10^{0.45} \times 8^{2.5} = 510\)
Question 2Medium9 marks
A biologist records the number of bacteria, \(P\), in a culture \(t\) hours after the start of an experiment, for \(0 \le t \le 10\). A graph of \(\log_{10} P\) against \(t\) is a straight line passing through the points \((0, 2.1)\) and \((10, 3.6)\).
(a) Explain why this suggests a model of the form \(P = ab^t\), where \(a\) and \(b\) are constants.[2]
(b) Find the values of \(a\) and \(b\), giving your answers to 3 significant figures.[3]
(c) Interpret the value of \(b\) in the context of the model.[1]
(d) Use the model to estimate the number of bacteria after 20 hours, and comment on the reliability of this estimate.[3]
Show the answer and mark scheme
(a)Answer: \(\log_{10} P = \log_{10} a + t\log_{10} b\), which is linear in \(t\) with gradient \(\log_{10} b\) and intercept \(\log_{10} a\).
M1 for taking logs: \(\log_{10} P = \log_{10} a + t\log_{10} b\)
A1 for identifying this as a straight line \(Y = mt + c\) with \(m = \log_{10} b\), \(c = \log_{10} a\)
Worked solution: If \(P = ab^t\), then \(\log_{10} P = \log_{10} a + t\log_{10} b\). This is of the form \(Y = c + mt\), so a graph of \(\log_{10} P\) against \(t\) is a straight line with gradient \(\log_{10} b\) and intercept \(\log_{10} a\).
(b)Answer: \(a = 126\), \(b = 1.41\)
M1 for gradient \(\frac{3.6 - 2.1}{10} = 0.15\)
M1 for \(a = 10^{2.1}\) or \(b = 10^{0.15}\)
A1 for \(a = 126\) and \(b = 1.41\)
Worked solution: Intercept \(\log_{10} a = 2.1\), so \(a = 10^{2.1} = 126\). Gradient \(\log_{10} b = \frac{3.6 - 2.1}{10} = 0.15\), so \(b = 10^{0.15} = 1.41\).
(c)Answer: The number of bacteria is multiplied by about 1.41 each hour (it increases by about 41% per hour).
B1 for the number of bacteria increases by about 41% each hour (or is multiplied by 1.41 every hour)
Worked solution: Each hour \(P\) is multiplied by \(b = 1.41\): the population grows by about 41% per hour.
(d)Answer: About 126 000; unreliable because it extrapolates beyond the data (\(t \le 10\)) and growth will be limited by nutrients or space.
M1 for \(P = 10^{2.1 + 0.15 \times 20}\) or \(126 \times 1.41^{20}\)
A1 for about 126 000 (see notes for the accepted range)
B1 for a comment: unreliable, since \(t = 20\) is outside the range of the data (extrapolation) and/or exponential growth cannot continue (limited food or space)
Worked solution: \(\log_{10} P = 2.1 + 0.15 \times 20 = 5.1\), so \(P = 10^{5.1} \approx 126\,000\). This is unreliable: the data only cover \(0 \le t \le 10\), so this is extrapolation, and in reality the growth of bacteria slows when nutrients or space run out.
Question 3Hard10 marks
The number of bacteria in a culture, \(N\), is modelled by \(N = kb^{t}\), where \(t\) is the time in hours and \(k\) and \(b\) are constants. The diagram shows the straight line graph of \(\ln N\) against \(t\), which passes through \(A(0, 4.8)\) and \(B(8, 8.48)\).
[object Object]
(a) Show that \(\ln N = \ln k + t\ln b\).[2]
(b) Find the values of \(k\) and \(b\), giving \(k\) to 3 significant figures and \(b\) to 4 significant figures.[4]
(c) Interpret the value of \(b\) in the context of the model.[1]
(d) Find the value of \(t\) when the model predicts that \(N = 44{,}000\). Give your answer to 3 significant figures.[2]
(e) Give one reason why the model may not be reliable for predicting the value of \(N\) far into the future.[1]
Show the answer and mark scheme
(a)Answer: Shown
M1 for \(\ln N = \ln k + \ln b^{t}\)
A1* for \(\ln b^{t} = t\ln b\)
Worked solution: \(\ln N = \ln(kb^{t}) = \ln k + \ln b^{t} = \ln k + t\ln b\)
(b)Answer: \(k = 122, \ b = 1.584\)
M1 for \(\ln k = 4.8\)
A1 for \(k = 122\)
M1 for \(\ln b = \) gradient \(= \frac{8.48 - 4.8}{8} = 0.46\)
A1 for \(b = 1.584\)
Worked solution: The line meets the vertical axis at \((0, 4.8)\), so \(\ln k = 4.8 \Rightarrow k = \mathrm{e}^{4.8} = 122\). Gradient \(= \frac{8.48 - 4.8}{8} = 0.46 = \ln b \Rightarrow b = \mathrm{e}^{0.46} = 1.584\).
(c)Answer: The number of bacteria in a culture increases by about 58% each hour.
B1 for the number is multiplied by 1.584 each hour (an increase of about 58% per hour)
Worked solution: Each hour, \(N\) is multiplied by \(1.584\), i.e. it grows by about 58% per hour.
(d)Answer: \(t = 12.8\)
M1 for \(\ln 44{,}000 = 4.8 + 0.46t\)
A1 for awrt \(12.8\)
Worked solution: \(t = \frac{\ln 44{,}000 - 4.8}{0.46} = 12.8\).
(e)Answer: The model predicts unlimited exponential growth, which cannot continue indefinitely (e.g. limited food or space); predictions far outside the data range are extrapolation.
B1 for a sensible reason, e.g. exponential growth cannot continue for ever / limited resources / extrapolation beyond the range of the data
Worked solution: The model assumes growth by the same factor every hour for ever. In practice growth is limited (food and space run out), and using the model far beyond the data is extrapolation.