| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve by showing reduces to polynomial |
| Difficulty | Standard +0.3 This is a straightforward logarithm manipulation question requiring standard laws (power rule, addition rule) to convert to polynomial form, then applying discriminant condition for one root. The algebraic steps are routine and well-signposted by the question structure, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_2 x^2 \left(= \log_2(kx-1) + 3\right)\) | B1 | Using \(a\log b = \log(b^a)\) |
| \(\log_2\left(\dfrac{x^2}{kx-1}\right) = 3\) | M1\* | Re-arranging and correctly combining both log terms; or re-write 3 as \(\log_2 8\) |
| \(\dfrac{x^2}{kx-1} = 2^3\), so \(x^2 = 8(kx-1)\) | Dep\*M1 | Correctly remove logs |
| \(x^2 - 8kx + 8 = 0\) | A1 | AG; must show sufficient working |
| Answer | Marks | Guidance |
|---|---|---|
| \(b^2 - 4ac = 0 \Rightarrow (-8k)^2 - 4(1)(8) = 0\) | M1 | Use of \(b^2 - 4ac = 0\); or state equation must be of the form \((x+p)^2 = 0\) with \(p^2 = 8\) |
| \(k = (\pm)\dfrac{1}{\sqrt{2}}\) | A1 | oe exact |
| \(k = \dfrac{1}{\sqrt{2}} \Rightarrow x = 2\sqrt{2}\) | A1 | BC oe exact; so \(x = (\pm)2\sqrt{2}\) |
| \(k = -\dfrac{1}{\sqrt{2}} \Rightarrow x = -2\sqrt{2}\) and as \(\log_2 x\) is only defined for \(x > 0\) so \(x \neq -2\sqrt{2}\) | A1 | BC oe statement for rejection of negative value of \(x\); reject \(x = -2\sqrt{2}\) with valid reason |
# Question 8:
## Part (a):
$\log_2 x^2 \left(= \log_2(kx-1) + 3\right)$ | **B1** | Using $a\log b = \log(b^a)$
$\log_2\left(\dfrac{x^2}{kx-1}\right) = 3$ | **M1\*** | Re-arranging and correctly combining both log terms; or re-write 3 as $\log_2 8$
$\dfrac{x^2}{kx-1} = 2^3$, so $x^2 = 8(kx-1)$ | **Dep\*M1** | Correctly remove logs
$x^2 - 8kx + 8 = 0$ | **A1** | AG; must show sufficient working
## Part (b):
$b^2 - 4ac = 0 \Rightarrow (-8k)^2 - 4(1)(8) = 0$ | **M1** | Use of $b^2 - 4ac = 0$; or state equation must be of the form $(x+p)^2 = 0$ with $p^2 = 8$
$k = (\pm)\dfrac{1}{\sqrt{2}}$ | **A1** | oe exact
$k = \dfrac{1}{\sqrt{2}} \Rightarrow x = 2\sqrt{2}$ | **A1** | BC oe exact; so $x = (\pm)2\sqrt{2}$
$k = -\dfrac{1}{\sqrt{2}} \Rightarrow x = -2\sqrt{2}$ and as $\log_2 x$ is only defined for $x > 0$ so $x \neq -2\sqrt{2}$ | **A1** | BC oe statement for rejection of negative value of $x$; reject $x = -2\sqrt{2}$ with valid reason8
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $2 \log _ { 2 } x = \log _ { 2 } ( k x - 1 ) + 3$, where $k$ is a constant, can be expressed in the form $x ^ { 2 } - 8 k x + 8 = 0$.
\item Given that the equation $2 \log _ { 2 } x = \log _ { 2 } ( k x - 1 ) + 3$ has only one real root, find the value of this root.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q8 [8]}}