| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Two unrelated log parts: one non-log algebraic part |
| Difficulty | Moderate -0.3 Part (a) requires applying a standard log law (ln a - ln b = ln(a/b)) and exponentiating, which is routine AS-level technique. Part (b) involves substituting into a linear equation and solving a simple equation involving e³, requiring algebraic manipulation but no novel insight. This is slightly easier than average due to being a straightforward application of standard techniques with clear steps. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks |
|---|---|
| 8(a) | Obtains e3 |
| Answer | Marks | Guidance |
|---|---|---|
| Accept AWRT 20.1 | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| appropriately | 1.1a | M1 |
| Obtains x = e3y | 1.1b | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 8(b) | Obtains a correct equation in x or |
| Answer | Marks | Guidance |
|---|---|---|
| does not involve logarithms | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1+e3 1+e3 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1+e3 1+e3 | 1.1b | A1 |
| Subtotal | 3 | |
| Question 8 Total | 6 | |
| Q | Marking instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Obtains e3
May be seen anywhere
Accept AWRT 20.1 | 1.1b | B1 | ln x – ln y = 3
x
ln = 3
y
x
= e3
y
x = e3y
Uses a law of logarithms
appropriately
or
Uses a rules of indices
appropriately | 1.1a | M1
Obtains x = e3y | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b) ---
8(b) | Obtains a correct equation in x or
y only
FT their answer to (a) provided it
does not involve logarithms | 1.1a | M1 | e3y + y = 10
10
y =
1+e3
10e3
x =
1+e3
10e3 10
Obtains x = or y =
1+e3 1+e3 | 1.1b | A1
10e3 10
Obtains x = and y =
1+e3 1+e3 | 1.1b | A1
Subtotal | 3
Question 8 Total | 6
Q | Marking instructions | AO | Marks | Typical solutionIt is given that
$$\ln x - \ln y = 3$$
\begin{enumerate}[label=(\alph*)]
\item Express $x$ in terms of $y$ in a form not involving logarithms.
[3 marks]
\item Given also that
$$x + y = 10$$
find the exact value of $y$ and the exact value of $x$
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2024 Q8 [6]}}