| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Express log in terms of given variables |
| Difficulty | Moderate -0.3 This is a straightforward application of logarithm laws (power rule, product rule) with minimal problem-solving required. Parts (a) and (b) are routine manipulations (256=16², 100=25×4), while part (c) requires recognizing 80=16×5 and 3.2=16/5, then applying the product of logs formula—slightly more involved but still standard practice for AS level. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2p\) | B1 | o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\log_a 100 = \log_a 4 + \log_a 25\) | M1 | Uses laws of logs to write \(\log_a 100\) correctly as a sum of logs. e.g. \(\log_a 100 = \log_a 4 + \log_a a^q\); e.g. \(\log_a 100 = 2\log_a 2 + 2\log_a 5\); e.g. \(\log_a 100 = \frac{1}{2}\log_a 16 + \frac{1}{2}\log_a 625\); e.g. \(\log_a 100 = \log_a 50 + \log_a 2\) |
| \(\log_a 16^{\frac{1}{2}} + \log_a 25 = \frac{1}{2}p + q\) | A1 | \(\frac{1}{2}p + q\) o.e. Withhold if incorrect log work seen e.g. \(\log_a 100 = \log_a 4 \times \log_a 25\) or \(\log_a 100 = 4\log_a 25\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| e.g. \(\log_a 80 \times \log_a 3.2 = (\log_a 16 + \log_a 5)\times(\log_a 16 - \log_a 5)\) | M1 | Uses both addition and subtraction laws to write full expression in terms of: \(\log_a 16\ (=p)\); \(\log_a 4\ \left(=\frac{p}{2}\right)\); \(\log_a 2\ \left(=\frac{p}{4}\right)\); \(\log_a 5\ \left(=\frac{q}{2}\right)\); \(\log_a 25\ (=q)\). Must proceed to either an integer value (e.g. \(\log_a 5\)) or expression in \(p\) or \(q\). Do not penalise omission of base \(a\) |
| \(\left(p+\frac{1}{2}q\right)\times\left(p-\frac{1}{2}q\right)\) or \(p^2-\frac{1}{4}q^2\) | A1 | o.e. Does not need to be simplified. Correct answer scores full marks but withhold if incorrect log work seen |
## Question 9:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $2p$ | B1 | o.e. |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_a 100 = \log_a 4 + \log_a 25$ | M1 | Uses laws of logs to write $\log_a 100$ correctly as a sum of logs. e.g. $\log_a 100 = \log_a 4 + \log_a a^q$; e.g. $\log_a 100 = 2\log_a 2 + 2\log_a 5$; e.g. $\log_a 100 = \frac{1}{2}\log_a 16 + \frac{1}{2}\log_a 625$; e.g. $\log_a 100 = \log_a 50 + \log_a 2$ |
| $\log_a 16^{\frac{1}{2}} + \log_a 25 = \frac{1}{2}p + q$ | A1 | $\frac{1}{2}p + q$ o.e. Withhold if incorrect log work seen e.g. $\log_a 100 = \log_a 4 \times \log_a 25$ or $\log_a 100 = 4\log_a 25$ |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. $\log_a 80 \times \log_a 3.2 = (\log_a 16 + \log_a 5)\times(\log_a 16 - \log_a 5)$ | M1 | Uses both addition and subtraction laws to write full expression in terms of: $\log_a 16\ (=p)$; $\log_a 4\ \left(=\frac{p}{2}\right)$; $\log_a 2\ \left(=\frac{p}{4}\right)$; $\log_a 5\ \left(=\frac{q}{2}\right)$; $\log_a 25\ (=q)$. Must proceed to either an integer value (e.g. $\log_a 5$) or expression in $p$ or $q$. Do not penalise omission of base $a$ |
| $\left(p+\frac{1}{2}q\right)\times\left(p-\frac{1}{2}q\right)$ or $p^2-\frac{1}{4}q^2$ | A1 | o.e. Does not need to be simplified. Correct answer scores full marks but withhold if incorrect log work seen |
---9.
$$\begin{aligned}
p & = \log _ { a } 16 \\
q & = \log _ { a } 25
\end{aligned}$$
where $a$ is a constant.\\
Find in terms of $p$ and/or $q$,
\begin{enumerate}[label=(\alph*)]
\item $\log _ { a } 256$
\item $\log _ { a } 100$
\item $\log _ { a } 80 \times \log _ { a } 3.2$
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2024 Q9 [5]}}