| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Two unrelated log/algebra parts - simplify/express then solve |
| Difficulty | Moderate -0.8 Part (a) is a straightforward application of logarithm laws (addition, power rule) requiring 4 routine steps to reach y = 196√a. Part (b) tests understanding that logarithms are only defined for positive arguments, so x = -3/2 is extraneous—a common conceptual check but requiring only recognition rather than problem-solving. Both parts are below-average difficulty, being standard textbook exercises on logarithm manipulation and domain restrictions. |
| Spec | 1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | Uses nlog x=log xn correctly | |
| a a | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| a a a y | AO1.1a | M1 |
| Obtains a | AO1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| correct form. | AO1.1b | A1 |
| Answer | Marks |
|---|---|
| 7(b) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| a 2 | AO2.3 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 5 | |
| Q | Marking Instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Uses nlog x=log xn correctly
a a | AO1.1a | M1 | 1
log y =2log 7+log 4+
a a a 2
1
⇒log y =log 72 +log 4+
a a a 2
1
=log ( 49×4 )+
a 2
1
=log 196+ log a
a 2 a
=log 196+log a
a a
=log 196 a
a
∴y =196 a
Uses log x+log y =log xy or
a a a
x
log x−log y =log correctly
a a a y | AO1.1a | M1
Obtains a | AO1.1b | B1
Obtains correct answer in any
correct form. | AO1.1b | A1
--- 7(b) ---
7(b) | 3
Explains that − should be
2
rejected as it is not possible to
3
evaluate log −
a 2 | AO2.3 | E1 | 3
− should be rejected as it is not
2
3
possible to evaluate log −
a 2
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution\begin{enumerate}[label=(\alph*)]
\item Given that $\log_a y = 2\log_a 7 + \log_a 4 + \frac{1}{2}$, find $y$ in terms of $a$.
[4 marks]
\item When asked to solve the equation
$$2\log_a x = \log_a 9 - \log_a 4$$
a student gives the following solution:
$2\log_a x = \log_a 9 - \log_a 4$
$\Rightarrow 2\log_a x = \log_a \frac{9}{4}$
$\Rightarrow \log_a x^2 = \log_a \frac{9}{4}$
$\Rightarrow x^2 = \frac{9}{4}$
$\therefore x = \frac{3}{2}$ or $-\frac{3}{2}$
Explain what is wrong with the student's solution.
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2018 Q7 [5]}}