Edexcel A level Maths (9MA0) · Pure mathematics › Exponentials and logarithms
Practise Using log graphs to estimate parameters. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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 \(-1.5\) that meets the vertical axis at \((0, -0.4)\). 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: \(n = -1.5, \ a = 0.398\)
B1 for \(n = -1.5\)
M1 for \(\log_{10} a = -0.4\)
A1 for \(a = 0.398\)
Worked solution: Comparing with \(Y = mX + c\): the gradient gives \(n = -1.5\) and the intercept gives \(\log_{10} a = -0.4 \Rightarrow a = 10^{-0.4} = 0.398\).
(c)Answer: \(y = 0.0176\)
M1 for \(0.398 \times 8^{-1.5}\) (or \(\log_{10} y = -0.4 - 1.5\log_{10} 8\))
A1 for awrt \(0.0176\)
Worked solution: \(y = 10^{-0.4} \times 8^{-1.5} = 0.0176\)
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 \(\log_{10} N\) against \(t\), which passes through \(A(0, 1.8)\) and \(B(10, 3.8)\).
[object Object]
(a) Show that \(\log_{10} N = \log_{10} k + t\log_{10} b\).[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 \(N = 100{,}000\). 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 \(\log_{10} N = \log_{10} k + \log_{10} b^{t}\)
A1* for \(\log_{10} b^{t} = t\log_{10} b\)
Worked solution: \(\log_{10} N = \log_{10}(kb^{t}) = \log_{10} k + \log_{10} b^{t} = \log_{10} k + t\log_{10} b\)
(b)Answer: \(k = 63.1, \ b = 1.585\)
M1 for \(\log_{10} k = 1.8\)
A1 for \(k = 63.1\)
M1 for \(\log_{10} b = \) gradient \(= \frac{3.8 - 1.8}{10} = 0.2\)
A1 for \(b = 1.585\)
Worked solution: The line meets the vertical axis at \((0, 1.8)\), so \(\log_{10} k = 1.8 \Rightarrow k = 10^{1.8} = 63.1\). Gradient \(= \frac{3.8 - 1.8}{10} = 0.2 = \log_{10} b \Rightarrow b = 10^{0.2} = 1.585\).
(c)Answer: The number of bacteria in a culture increases by about 58% each hour.
B1 for the number is multiplied by 1.585 each hour (an increase of about 58% per hour)
Worked solution: Each hour, \(N\) is multiplied by \(1.585\), i.e. it grows by about 58% per hour.
(d)Answer: \(t = 16.0\)
M1 for \(\log_{10} 100{,}000 = 1.8 + 0.2t\)
A1 for awrt \(16.0\)
Worked solution: \(t = \frac{\log_{10} 100{,}000 - 1.8}{0.2} = 16.0\).
(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.