| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| 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 logarithm manipulation question requiring standard laws (power rule, converting to exponential form) to reduce to a quadratic, then solving. Part (b) is a simple substitution. The steps are routine and well-practiced at A-level, making it slightly easier than average despite being a 'show that' question. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Step 1 | M1 | Use law of logarithm of a power: \(\log_3(2x+1) = 1 + \log_3(x-1)^2\) |
| Step 2 | B1 | Use \(\log_3 3 = 1\): \(\log_3(2x+1) = \log_3 3 + 2\log_3(x-1)\), giving \(\log_3\!\left(\frac{2x+1}{(x-1)^2}\right) = \log_3 3\) or \(\frac{2x+1}{(x-1)^2} = 3\). SC: for candidates scoring M0 B0 due to combining logs before dealing with coefficient 2 and confusing coefficients, allow \(\log_3(\ldots)=c\) leading to \((\ldots)=3^c\) B1. |
| Step 3 | A1 | Obtain \(3x^2 - 8x + 2 = 0\) or \(1.5x^2 - 4x + 1 = 0\). OE 3 terms only and \(= 0\) required. |
| Answer | Marks | Guidance |
|---|---|---|
| Step 1 | M1 | Solve 3-term quadratic from part 3(a) or restart to find \(y\): \(y = \frac{4\pm\sqrt{10}}{6}\) or \(y = 1.1937\ldots\) or \(y = 0.1396\ldots\) (i.e. \(x = 2.3874\) or \(x = 0.2792\)). May solve for \(x\) but must find \(y = \frac{x}{2}\) to gain M1. |
| Step 2 | A1 | Obtain answer \(1.19\). CAO. 2 dp required. |
## Question 3(a):
**Step 1** | M1 | Use law of logarithm of a power: $\log_3(2x+1) = 1 + \log_3(x-1)^2$
**Step 2** | B1 | Use $\log_3 3 = 1$: $\log_3(2x+1) = \log_3 3 + 2\log_3(x-1)$, giving $\log_3\!\left(\frac{2x+1}{(x-1)^2}\right) = \log_3 3$ or $\frac{2x+1}{(x-1)^2} = 3$. SC: for candidates scoring M0 B0 due to combining logs before dealing with coefficient 2 and confusing coefficients, allow $\log_3(\ldots)=c$ leading to $(\ldots)=3^c$ **B1**.
**Step 3** | A1 | Obtain $3x^2 - 8x + 2 = 0$ or $1.5x^2 - 4x + 1 = 0$. OE 3 terms only and $= 0$ required.
**Total: 3 marks**
---
## Question 3(b):
**Step 1** | M1 | Solve 3-term quadratic from part **3(a)** or restart to find $y$: $y = \frac{4\pm\sqrt{10}}{6}$ or $y = 1.1937\ldots$ or $y = 0.1396\ldots$ (i.e. $x = 2.3874$ or $x = 0.2792$). May solve for $x$ but must find $y = \frac{x}{2}$ to gain M1.
**Step 2** | A1 | Obtain answer $1.19$. CAO. 2 dp required.
**Total: 2 marks**
---3
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\log _ { 3 } ( 2 x + 1 ) = 1 + 2 \log _ { 3 } ( x - 1 )$ can be written as a quadratic equation in $x$.
\item Hence solve the equation $\log _ { 3 } ( 4 y + 1 ) = 1 + 2 \log _ { 3 } ( 2 y - 1 )$, giving your answer correct to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q3 [5]}}