| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve by showing reduces to polynomial |
| Difficulty | Moderate -0.3 This is a straightforward substitution question requiring recognition that 16^x = (2^4)^x = (2^2x)^2 = y^2 and 2^(2x+3) = 8·2^2x = 8y, followed by solving a simple quadratic and rejecting the negative solution. While it requires multiple steps and logarithm manipulation, these are standard AS-level techniques with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks |
|---|---|
| 8(a) | Uses rules of indices to express |
| Answer | Marks | Guidance |
|---|---|---|
| (42)x, not just 16x = (22x)2 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| convinced | 2.1 | R1 |
| Subtotal | 2 |
| Answer | Marks |
|---|---|
| 8(b) | Solves the equation in terms of 22x |
| Answer | Marks | Guidance |
|---|---|---|
| 22x = –1) | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| have a square root) | 2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| evidence for M1 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | 2.1 | R1 |
| Subtotal | 4 | |
| Question Total | 6 | |
| Q | Marking Instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Uses rules of indices to express
2(2x + 3) in terms of y
or
16x in terms of y, must see an
intermediary step, either (24)x or
(42)x, not just 16x = (22x)2 | 1.1a | M1 | 2(2x + 3) = 22x × 23 = 8y
16x = (24)x= (22x)2 = y2
y2 – 8y – 9 = 0
Completes rigorous solution
expressing both terms in terms of y
Condone working backwards from
quadratic in y to both terms in x. Be
convinced | 2.1 | R1
Subtotal | 2
--- 8(b) ---
8(b) | Solves the equation in terms of 22x
(PI by seeing either 22x = 9 or
22x = –1) | 1.1a | M1 | y = 9 or y = –1
22x = 9 or 22x = –1
States that 22x = –1 has no (real)
solutions
(PI by negative numbers do not
have a square root) | 2.4 | E1 | 22x = –1 has no solutions
Uses logs to solve 22x = 9 OE
Or
Square roots first then correctly
takes logs
Accept log 9 or log 9 seen as
2 4
evidence for M1 | 1.1a | M1 | 2x = log 9
2
x = log 9
2
1
= log
2 2
1
2
= log 39
2
Completes rigorous solution. Must
include justification for 9 becoming
3 | 2.1 | R1
Subtotal | 4
Question Total | 6
Q | Marking Instructions | AO | Marks | Typical Solution\begin{enumerate}[label=(\alph*)]
\item Using $y = 2^{2x}$ as a substitution, show that
$$16^x - 2^{(2x+3)} - 9 = 0$$
can be written as
$$y^2 - 8y - 9 = 0$$
[2 marks]
\item Hence, show that the equation
$$16^x - 2^{(2x+3)} - 9 = 0$$
has $x = \log_2 3$ as its only solution.
Fully justify your answer.
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2020 Q8 [6]}}