| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Model with logarithmic relationship: find constant or predict value |
| Difficulty | Moderate -0.3 This is a straightforward application of logarithm laws requiring students to (a) substitute values and solve for a constant, (b) rearrange using log laws to make t the subject, and (c) substitute and calculate. While it involves multiple steps and logarithm manipulation, each step follows standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(10=\log_a 8-\log_a 4 \Rightarrow \log_a 2=10\) | M1 | Substitutes \(t=3\), \(w=10\); achieves \(\log_a 2=10\) or e.g. \(\log_a\tfrac{8}{4}=10\) correctly |
| \(a^{10}=2\) | M1 | Correctly removes the log to obtain \(a^{10}=2\) |
| \(a=2^{1/10}=1.07177\ldots\) | A1* | Fully correct proof; \(a=2^{1/10}\) or \(a=\sqrt[10]{2}\) or \(a=\sqrt[10]{\tfrac{8}{4}}\); obtains awrt 1.072; allow 1.0718 (rounded) or 1.0717 (truncated) |
| Answer | Marks |
|---|---|
| Working | Marks |
| \(10=\log_a8-\log_a4\Rightarrow\dfrac{\log_a8}{\log_a4}=10\Rightarrow\log_a2=10\Rightarrow a^{10}=2\Rightarrow a=\sqrt[10]{2}=1.072\) | M0M1A0 |
| \(10=\log_a8-\log_a4\Rightarrow\dfrac{8\log a}{4\log a}=10\Rightarrow 2\log a=10\Rightarrow\log a^2=10\Rightarrow a^{10}=2\Rightarrow a=\sqrt[10]{2}=1.072\) | 0 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w=\log_{1.072}(t+5)-\log_{1.072}4 \Rightarrow w=\log_{1.072}\!\left(\dfrac{t+5}{4}\right)\) | M1 | Combines logs correctly |
| \(\dfrac{t+5}{4}=1.072^w \Rightarrow t=\ldots\) | M1 | Removes log correctly |
| \(t=4\times1.072^w-5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w+\log_{1.072}4=\log_{1.072}(t+5)\) | M1 | |
| \(t+5=1.072^{w+\log_{1.072}4}\) | M1 | |
| \(t=1.072^{w+\log_{1.072}4}-5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=4\times1.072^{15}-5=\ldots\) | M1 | Substitutes \(w=15\) into their expression from (b) |
| awrt \(6.35\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w = \log_{1.072}\left(\frac{t+5}{4}\right)\) | M1 | Applies subtraction law for logs |
| \(1.072^w = f(t)\) and makes \(t\) the subject | M1 | Writes in exponential form and rearranges |
| \(t = 4 \times 1.072^w - 5\) | A1 | Correct equation |
| Alternative: \(\log_{1.072}(t+5)\) as subject | M1 | Rearranges correctly |
| \(f(t) = 1.072^{g(w)}\) | M1 | Removes logs from lhs |
| \(t = 1.072^{w + \log_{1.072}4} - 5\) | A1 | Correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(w = 15\) into equation from (b) | M1 | Or uses \(w = \log_{1.072}(t+5) - \log_{1.072}4\) |
| \(t \approx 6.35\) (months) | A1 | If full accuracy used for \(a\), answer is \(6.3137...\) and scores A0 |
# Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $10=\log_a 8-\log_a 4 \Rightarrow \log_a 2=10$ | M1 | Substitutes $t=3$, $w=10$; achieves $\log_a 2=10$ or e.g. $\log_a\tfrac{8}{4}=10$ correctly |
| $a^{10}=2$ | M1 | Correctly removes the log to obtain $a^{10}=2$ |
| $a=2^{1/10}=1.07177\ldots$ | A1* | Fully correct proof; $a=2^{1/10}$ or $a=\sqrt[10]{2}$ or $a=\sqrt[10]{\tfrac{8}{4}}$; obtains awrt 1.072; allow 1.0718 (rounded) or 1.0717 (truncated) |
**False solutions (no marks):**
| Working | Marks |
|---|---|
| $10=\log_a8-\log_a4\Rightarrow\dfrac{\log_a8}{\log_a4}=10\Rightarrow\log_a2=10\Rightarrow a^{10}=2\Rightarrow a=\sqrt[10]{2}=1.072$ | M0M1A0 |
| $10=\log_a8-\log_a4\Rightarrow\dfrac{8\log a}{4\log a}=10\Rightarrow 2\log a=10\Rightarrow\log a^2=10\Rightarrow a^{10}=2\Rightarrow a=\sqrt[10]{2}=1.072$ | 0 marks |
# Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w=\log_{1.072}(t+5)-\log_{1.072}4 \Rightarrow w=\log_{1.072}\!\left(\dfrac{t+5}{4}\right)$ | M1 | Combines logs correctly |
| $\dfrac{t+5}{4}=1.072^w \Rightarrow t=\ldots$ | M1 | Removes log correctly |
| $t=4\times1.072^w-5$ | A1 | |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w+\log_{1.072}4=\log_{1.072}(t+5)$ | M1 | |
| $t+5=1.072^{w+\log_{1.072}4}$ | M1 | |
| $t=1.072^{w+\log_{1.072}4}-5$ | A1 | |
# Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=4\times1.072^{15}-5=\ldots$ | M1 | Substitutes $w=15$ into their expression from (b) |
| awrt $6.35$ | A1 | |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \log_{1.072}\left(\frac{t+5}{4}\right)$ | M1 | Applies subtraction law for logs |
| $1.072^w = f(t)$ and makes $t$ the subject | M1 | Writes in exponential form and rearranges |
| $t = 4 \times 1.072^w - 5$ | A1 | Correct equation |
| **Alternative:** $\log_{1.072}(t+5)$ as subject | M1 | Rearranges correctly |
| $f(t) = 1.072^{g(w)}$ | M1 | Removes logs from lhs |
| $t = 1.072^{w + \log_{1.072}4} - 5$ | A1 | Correct equation |
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $w = 15$ into equation from (b) | M1 | Or uses $w = \log_{1.072}(t+5) - \log_{1.072}4$ |
| $t \approx 6.35$ (months) | A1 | If full accuracy used for $a$, answer is $6.3137...$ and scores A0 |
---\begin{enumerate}
\item The weight of a baby mammal is monitored over a 16 -month period.
\end{enumerate}
The weight of the mammal, $w \mathrm {~kg}$, is given by
$$w = \log _ { a } ( t + 5 ) - \log _ { a } 4 \quad 2 \leqslant t \leqslant 18$$
where $t$ is the age of the mammal in months and $a$ is a constant.\\
Given that the weight of the mammal was 10 kg when $t = 3$\\
(a) show that $a = 1.072$ correct to 3 decimal places.
Using $a = 1.072$\\
(b) find an equation for $t$ in terms of $w$\\
(c) find the value of $t$ when $w = 15$, giving your answer to 3 significant figures.
\hfill \mbox{\textit{Edexcel P2 2022 Q4 [8]}}