Logarithms and logarithmic graphs - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 19 exam-style questions on Logarithms and logarithmic graphs for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

19 questions
2–9 marks each
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19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Given that
\[\ln y = 2 + 3\ln x, \quad x > 0\]
express \(y\) in terms of \(x\), giving your answer in a form that does not involve logarithms.

Model answer

\(\ln y = 2 + 3\ln x = \ln \mathrm{e}^2 + \ln x^3 = \ln\left(\mathrm{e}^2x^3\right)\)
So \(y = \mathrm{e}^2x^3\)

Explanation

Write each term as a single logarithm, \(2 = \ln \mathrm{e}^2\) and \(3\ln x = \ln x^3\), then use the addition law to get \(\ln y = \ln\left(\mathrm{e}^2x^3\right)\). Since \(\ln\) is a one-to-one function, \(y = \mathrm{e}^2x^3\). A common error is to exponentiate each term separately and write \(y = \mathrm{e}^2 + x^3\).

Mark scheme

1 mark
M1 Uses the power law: \(3\ln x = \ln x^3\); or a correct final answer of \(y = \mathrm{e}^2x^3\) with no working
An answer of \(y = \mathrm{e}^2x^3\) with no working scores 3 marks.
1 mark
M1 Deals with the constant: writes \(2 = \ln \mathrm{e}^2\) and combines, \(\ln y = \ln\left(\mathrm{e}^2x^3\right)\); or takes exponentials of both sides, \(y = \mathrm{e}^{2 + 3\ln x} = \mathrm{e}^2 \times \mathrm{e}^{3\ln x}\)
Not dependent on the first M mark. \(y = \mathrm{e}^2 + x^3\) (exponentiating each term separately) scores M0.
1 mark
A1 \(y = \mathrm{e}^2x^3\)
Accept \(y = x^3\mathrm{e}^2\). Do not accept \(y = 7.39x^3\) unless \(\mathrm{e}^2x^3\) is seen first.

Question 2

4 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Find the smallest integer \(n\) for which
\[3^n > 2^{n + 10}\]

Model answer

\(3^n > 2^{n + 10}\)
\(n\ln 3 > (n + 10)\ln 2\)
\(n\ln 3 - n\ln 2 > 10\ln 2\)
\(n\ln\dfrac{3}{2} > 10\ln 2\)
\(n > \dfrac{10\ln 2}{\ln 1.5} = 17.09\ldots\)
The smallest integer is \(n = 18\).

Explanation

Taking logarithms turns the powers into products, \(n\ln 3 > (n + 10)\ln 2\), which is linear in \(n\). Since \(\ln 1.5 > 0\), dividing keeps the inequality the same way round, giving \(n > 17.09\), so the smallest integer is 18. Common errors are missing the brackets around \(n + 10\) and rounding 17.09 down to 17.

Mark scheme

1 mark
M1 Takes logarithms of both sides correctly: \(n\log 3 > (n + 10)\log 2\). An answer with no working scores 0 marks
An answer of \(n = 18\) with no working scores 0 marks: the question demands working. Allow "=" for the method marks.
1 mark
M1 (dM1) Collects the terms in \(n\): \(n(\log 3 - \log 2) > 10\log 2\), i.e. \(n > \dfrac{10\log 2}{\log 1.5}\)
The inequality does not reverse, as \(\log 3 - \log 2 > 0\).
1 mark
A1 \(n >\) awrt 17.1, i.e. in [17.05, 17.15)
The value is 17.095...
1 mark
A1 \(n = 18\)
cao. Do not accept \(n = 17\) or \(n > 17.1\) as the final answer.

Question 3

5 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Solve the simultaneous equations
\[\log_2 x + \log_2 y = 7\]
\[\log_2 x - \log_4 y = 1\]
where \(x > 0\) and \(y > 0\).

[You may use the change of base formula: \(\log_a x = \dfrac{\log_b x}{\log_b a}\)]

Model answer

\(\log_4 y = \dfrac{\log_2 y}{\log_2 4} = \dfrac{1}{2}\log_2 y\)
Let \(p = \log_2 x\) and \(q = \log_2 y\):
\(p + q = 7\)
\(p - \tfrac{1}{2}q = 1\)
Subtracting: \(\tfrac{3}{2}q = 6 \Rightarrow q = 4\), so \(p = 3\)
\(x = 2^3 = 8\) and \(y = 2^4 = 16\)

Explanation

Change \(\log_4 y\) to base 2 (it is \(\tfrac{1}{2}\log_2 y\), because \(\log_2 4 = 2\)); then both equations are linear in \(\log_2 x\) and \(\log_2 y\). Solving gives \(\log_2 x = 3\) and \(\log_2 y = 4\), so \(x = 8\) and \(y = 16\). A common error is to treat \(\log_4 y\) as \(2\log_2 y\).

Mark scheme

1 mark
M1 Changes the base of \(\log_4 y\): \(\log_4 y = \dfrac{\log_2 y}{\log_2 4} = \dfrac{1}{2}\log_2 y\). An answer with no working scores 0 marks
An answer of \(x = 8\), \(y = 16\) with no working scores 0 marks: the question demands working. Alternative method: \(xy = 2^7 = 128\) and \(\dfrac{x}{\sqrt{y}} = 2\) (from \(\log_2 x - \log_2 \sqrt{y} = 1\)); the M1 is for \(\log_4 y = \log_2 \sqrt{y}\).
1 mark
M1 Solves the pair of linear equations in \(\log_2 x\) and \(\log_2 y\): \(p + q = 7\), \(p - \tfrac{1}{2}q = 1\)
Not dependent on the first M mark. In the alternative, eliminates \(x\) or \(y\), e.g. \(2\sqrt{y} \times y = 128\).
1 mark
A1 \(\log_2 x = 3\) and \(\log_2 y = 4\)
Both needed. In the alternative, award for \(y^{3/2} = 64\) or \(x^3 = 512\).
1 mark
A1 \(x = 8\)
Depends on the second M mark.
1 mark
A1 \(y = 16\)
Depends on the second M mark.

Question 4

7 marks

A baked potato is taken out of an oven and left to cool in a kitchen. The temperature, \(\theta\) °C, at the centre of the potato \(t\) minutes after it is taken out of the oven is modelled by
\[\theta = 20 + A\mathrm{e}^{-kt}\]
where \(A\) and \(k\) are positive constants.

A graph of \(\ln(\theta - 20)\) against \(t\) is a straight line passing through the points (5, 3.98) and (15, 3.18).

Find \(k\) and \(A\), giving \(A\) correct to 3 significant figures.

Find, according to the model, the temperature at the centre of the potato when it is taken out of the oven.

The potato is cool enough to eat when the temperature at its centre has fallen to 60 °C. Find, according to the model, how long after it is taken out of the oven this happens. Give your answer in minutes, correct to 3 significant figures.

Model answer

\(\theta - 20 = A\mathrm{e}^{-kt} \Rightarrow \ln(\theta - 20) = \ln A - kt\)
This is a straight line with gradient \(-k\) and intercept \(\ln A\).
Gradient \(= \dfrac{3.18 - 3.98}{15 - 5} = -0.08\), so \(k = 0.08\)
Intercept: \(\ln A = 3.98 + 0.08 \times 5 = 4.38\), so \(A = \mathrm{e}^{4.38} = 79.8\) (3 s.f.)
When \(t = 0\): \(\theta = 20 + A = 99.8\) °C
When \(\theta = 60\): \(\ln 40 = 4.38 - 0.08t\)
\(t = \dfrac{4.38 - \ln 40}{0.08} = 8.64\) minutes (3 s.f.)

Explanation

Subtracting 20 and taking natural logarithms gives \(\ln(\theta - 20) = \ln A - kt\), so the gradient of the line is \(-k\) and its intercept is \(\ln A\): \(k = 0.08\) and \(A = \mathrm{e}^{4.38} = 79.8\). At the start the temperature is \(20 + A = 99.8\) °C, and \(\theta = 60\) gives \(\ln 40 = 4.38 - 0.08t\), so \(t = 8.64\) minutes. Common errors are taking logarithms of each term of \(20 + A\mathrm{e}^{-kt}\) and forgetting to add the 20 for the initial temperature.

Mark scheme

1 mark
M1 Writes \(\theta - 20 = A\mathrm{e}^{-kt}\) and takes natural logarithms, \(\ln(\theta - 20) = \ln A - kt\), matching it to the line (gradient \(-k\), intercept \(\ln A\)); or correct final answers of \(k = 0.08\), \(A = 79.8\), 99.8 °C and 8.64 minutes with no working (special case: 4 marks)
Correct answers of \(k = 0.08\), \(A = 79.8\), 99.8 °C and 8.64 minutes with no working score 4 marks (special case: 1 mark for each; accept each within the range given below). Taking logarithms of each term, e.g. \(\ln\theta = \ln 20 + \ln A - kt\), scores M0.
1 mark
A1 \(k = 0.08\)
From the gradient \(\dfrac{3.18 - 3.98}{15 - 5} = -0.08\). \(k = -0.08\) scores A0 (\(k\) is positive and the gradient is \(-k\)).
1 mark
M1 Finds the intercept of the line, \(3.98 + 0.08 \times 5 = 4.38\), and uses \(\ln A = \) their intercept to find \(A = \mathrm{e}^{4.38}\)
Accept substituting a point into \(\ln(\theta - 20) = \ln A - 0.08t\), e.g. \(3.98 = \ln A - 0.4\). Using \(A = 4.38\) or \(A = 10^{4.38}\) scores M0.
1 mark
A1 \(A =\) awrt 79.8, i.e. in [79.75, 79.85)
The value is \(\mathrm{e}^{4.38} = 79.838\ldots\)
1 mark
B1ft 99.8 °C (awrt 99.8, i.e. in [99.75, 99.85))
Follow through \(20 + \) their \(A\). Units are required (°C, degrees or degrees Celsius). 79.8 (forgetting to add 20) scores B0.
1 mark
M1 Sets \(\theta = 60\) and solves using logarithms: \(\ln 40 = 4.38 - 0.08t\), or \(A\mathrm{e}^{-0.08t} = 40\) with their \(A\) and \(k\)
Setting \(A\mathrm{e}^{-0.08t} = 60\) (forgetting the 20) scores M0.
1 mark
A1 \(t =\) awrt 8.64, i.e. in [8.635, 8.645)
The value is 8.639... from \(t = \dfrac{4.38 - \ln 40}{0.08}\). Accept 8.63 from using \(A = 79.8\). Units not required (the question asks for minutes).

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